UC-NRLF 


ON  THE 


THEORY  AND  CALCULATION 


CP 


BY 


MANSFIELD  MERRIMAN,  C.E.,  Ph.D., 

SHEFFIELD 


INSTRUCTOR    IN 

SCIEN 


THE 


NEW   YORK: 

D.  VAN  NOSTRAND,  PUBLISHER, 
23  MURRAY  AND  27  WARREN  STREET, 

1876. 


\ 


If' 


o 


PKEFACE, 

The  following  pages  are  divided  into  three 
chapters.  The  first  presents  by  way  of  intro 
duction  some  of  the  elementary  principles  of 
continuous  girders,  and  the  fundamental  ideas 
relating  to  the  calculation  of  strains.  The  sec 
ond  gives  the  theory  of  flexure  as  applied  to 
the  continuous  trjiss^  of  constant  cross  section, 
and  exhibits  it  in  formulae  (I)  to  (VI),  ready 
for  application  to  any  particular  case  ;  and  the 
third  gives  an  example  of  the  computation  of 
strains  in  a  continuous  truss  of  five  unequal 
spans,  with  some  useful  hints  concerning  the 
practical  building  of  such  bridges. 

The  theory  of  flexure  indicates  that,  by  the 
use  of  continuous  instead  of  single  span 
bridges,  a  saving  in  material  of  from  twenty  to 
forty  per  cent,  may  be  effected.  f It  is  easy  in 
deed  to  say  that  this  advantage  will  be  entirely 
swallowed  up  by  the  effect  of  changes  of  tem 
perature,  increased  labor  of  erection,  or  addi 
tional  cost  of  workmanship,  but  by  no  amount 
of  reasoning  can  such  disadvantages  be  esti 
mated.  Theory  indicates  a  large  saving, 


whether  or  not  it  can  be  realized,  may  only  be 
determined  by  trial.  Other  nations  have  built 
and  are  building  continuous  bridges,  and  their 
experience  has  not  yet  shown  that  the  system 
is  inferior  to  that  of  single  spans.  The  inter 
est  now  prevailing  among  American  engineers 
in  the  subject,  and  the  fact  that  at  some  recent 
bridge  let  tings  plans  have  been  offered  for  a 
continuous  structure,  seem  to  indicate  that  the 
system  will  also  be  tried  here. 

This  little  book  may  then  perhaps  be  of  val 
ue  to  bridge  engineers,  as  well  as  to  students 
in  general. 

M.  M. 

New  Haven,  Conn. ,  July  10,  1876. 


THEORY  AND  CALCULATION 


OP 


CONTINUOUS  BRIDGES. 


WHEN  a  straight  bridge  consists  of 
several  spans,  each  entirely  independent 
of  the  others,  it  is  said  to  be  composed 
of  simple  girders.  If,  on  the  other  hand, 
it  consists  of  a  single  truss  extending 
from  one  abutment  to  the  other  without 
any  disconnection  of  parts  over  the  piers 
it  is  called  a  continuous  girder.  A  load 
placed  upon  any  span  of  a  continuous 
beam  influences,  to  some  extent,  each  of 
the  other  spans,  and  hence  its  complete 
theory  is  much  more  complex  than  that 
of  the  simple  one.  This  very  complexity 
however  has  rendered  the  subject  an  at 
tractive  one  to  mathematicians,  who, 
pursuing  science  for  science's  sake,  have 


6 

investigated  the  laws  of  equilibrium 
which  govern  it.  These  laws  with  the 
many  beautiful  consequences  attending 
them  form  one  of  the  most  interesting 
chapters  of  mathematical  analysis,  and 
as  such  have  interest  and  value  inde 
pendent  of  their  application  in  engineer 
ing  art. 

It  is  the  object  of  the  present  paper  to 
present  in  as  simple  a  form  as  possible 
some  of  the  main  principles  and  laws 
most  needed  by  the  engineer,  and  to  il 
lustrate  their  application  as  fully  as 
space  will  permit  to  the  practical  de 
signing  of  continuous  bridges. 

CHAPTER  1. 

The  first  point  to  be  observed  in  con 
sidering  either  a  simple  or  continuous 
girder  is  that  all  the  exterior  forces 
which  act  upon  it  are  in  equilibrium. 
The  exterior  forces  embrace  the  weight 
of  the  girder  and  the  loads  upon  it  which 
act  downward,  and  the  pressures  or  re 
actions  of  the  supports  which  act  up- 


ward.  In  order  that  these  may  be  in 
equilibrium,  it  is  necessary  that  the  sum 
of  the  reactions  of  all  the  supports  must 
be  equal  to  the  total  weight  of  the  girder 
and  its  load. 

Thus,  if   a  simple  girder  of  uniform 
section  and  weight  rest  at  its  ends  upon 
two  supports,  the  reaction  of  each  sup 
port  will  be  one-half  the  weight.    Exactly 
in  the  center  between  the  two  supports 
or  abutments,  let   us  suppose  a  pier  to 
be  placed  just  touching,  but  not  pressing 
against  the  beam,  which,  at  that  point, 
has   a   deflection  below   a  straight  line 
joining   the    two  abutments.     Then  the 
condition  of  things  is  in  no  way  altered, 
for  the  weight  being  W,  each  abutment 
reacts  with  a  force  \  W,  while  the  pier 
bears  no  load.     Raise  now  the  pier  so  as 
to  lift  the   girder  above  the   line  of  de 
flection    and   it   receives   a   part  of  the 
weight   W,   while    Jie  reactions  of  the 
abutments   become   less  than   \  W.     If 
the  pier  be  raised  higher  and  higher,  it 
will    at  length   lift  the    girder   entirely 
from  the  abutments  and  bear  itself  the 


8 

whole  load  W.  In  every  position,  how 
ever,  the  sum  of  the  reactions  of  the 
three  supports  is  equal  to  the  total  load. 
For  example,  when  the  three  are  on  the 
same  level  it  may  be  shown  that  the  re 
action  of  each  abutment  is  -fs  W,  and 
that  of  the  pier  ii  W. 

This  illustration  shows  also  that  small 
differences  of  level  in  the  supports  occur 
ring  after  the  erection  of  a  bridge  cause 
large  variations  in  the  reactions  of  its 
supports  and  in  the  strains  in  its  several 
parts.  A  simple  girder  having  a  deflec 
tion  of  one  inch,  would,  if  raised  one 
and  three-fifth  inches  at  the  center,  be 
entirely  lifted  from  the  abutments.  In 
the  first  case  the  upper  fiber  would  be  in 
compression,  the  lower  in  tension;  in  the 
second  case,  the  upper  would  be  in  ten 
sion,  the  lower  in  compression.  If  the 
center  were  raised  only  one  inch,  the  re 
versal  would  be  only  partial,  the  upper 
fiber  becoming  subject  to  tension  for  a 
short  distance  on  each  side  of  the  mid 
dle.  This  fact  often  used  as  an  argu 
ment  against  continuous  bridges,  is  really 


9 

an  objection  only  when  the  piers  are 
liable  to  settle  after  erection.  Difference8 
of  level,  previously  existing,  do  not  act 
prejudicial  when  the  bridge  is  built  upon 
the  piers,  and  with  a  profile  correspond 
ing  to  them. 

The  mathematical  theory  of  the  con 
tinuous  girder  enables  its  reactions  and 
internal  strains  to  be  found  for  any  as 
sumed  levels  of  the  supports,  provided 
only  that  the  differences  of  level  are  very 
small  compared  wTith  the  length  of  the 
spans.     However  interesting  such  inves 
tigations  may  be  in  themselves,  they  are 
of  little  importance  in  practice,  since  it 
has  been  shown  that  when  all  the  points 
of    support  are  on   the   same  level,  the 
greatest  economy  of   material  results.* 
In  all  that  follows,  then,  we  shall  regard 
the  girder  as  resting  on  level  supports, 
or,  what  is  the  same  thing,  that  it  was 
built  with  a  profile  corresponding  to  that 
of  the  piers. 

The  loads  upon  a  bridge  and  the  reac- 

*  Weyrauch ;  Theorie  der  continuir  lichen  Trdger,  p.  129. 
Winkler  ;  Lehre  der  Elasticitdt,  p.  155. 


10 

tions  of  the  supports  are  external  forces. 
The  equilibrium  between  them  is  main 
tained  by  means  of  internal  forces,  which, 
in  a  framed  truss,  are  transmitted  longi 
tudinally  along  the  pieces  as  strains  of 
tension  and  compression.  When  all  the 
external  forces  are  known,  these  internal 
forces  or  strains  can  be  readily  found. 
This  very  important  point  we  shall  now 
proceed  to  illustrate. 

Fig.  1  represents  a  portion  of  a  con- 
Fig,  i, 


A 

Rl  R 


tinuous  girder;  the  first  span  on  the  left 
is  called  Zl5  which  also  represents  its 
length,  the  second  J2,  the  third  /3,  etc.  ; 
in  like  manner,-  the  supports  beginning 
on  the  left  are  designated  by  the  indices 
1,  2,  3,  etc.,  and  their  reactions  by  R,, 
R2,  R3,  etc.  Let  the  load  per  linear  unit 
be  iv,  supposed  uniformly  distributed, 
then  the  weight  of  the  first  span  will  be 
wl  of  the  second  wl  of  the  third  w  l 


11 


etc.     If  there  are  four  spans  the  total 


weight  will  be 


and  from  the  fundamental  idea  of  equili 
brium,  we  must  have  the  equation 


Each  of  the  reactions  is  then  a  fraction 
al  part  of  the  total  load,  and  by  methods 
hereafter  to  be  explained,  their  values 
may  be  readily  computed,  whatever  be 
the  number  of  spans.  Granting  for  the 
present  that  they  may  be  found,  let  us 
inquire  how  we  may  obtain  the  internal 
forces  or  strains  in  any  part  of  the  gir 
der. 

In  the  span  lz  let  a  vertical  plane  be 
passed,  cutting  the  beam  at  a  point 
whose  distance  from  the  support  3  is  x. 
All  the  internal  forces  acting  in  this  sec 
tion  may  be  considered  as  resolved  into 
two  components,  one  vertical  and  the 
other  horizontal.  The  sum  of  all  the 
vertical  components  is  a  force  which  pre 
vents  the  two  parts  of  the  beam  from 


12 

shearing  asunder,  and  is  called  the  shear 
for  that  section  ;  the  horizontal  compo 
nents  acting  in  parallel  planes  are  the 
resisting  strains  of  tension  and  compres 
sion  in  the  horizontal  fibers,  and  the  sum 
of  their  moments  with  reference  to  any 
point  in  the  section  is  called  the  moment 
of  resistance,  or  simply  the  moment  for 
that  section.  The  internal  strains  in  any 
section  are  thus  completely  represented 
by  the  shear  and  moment.  For  example, 
if  the  girder  in  Fig.  1  be  a  framed  truss 
of  which  Fig.  2  represents  the  span  13 

Fig.  2, 
A 


enlarged,  and  the  section  be  passed  cut 
ting  the  three  pieces  E  F,  F  e,  and  ef, 
the  vertical  components  of  the  chord 
strains  will  be  zero,  and  that  of  the  di 
agonal  strain  e  F  will  be  the  shear. 
Hence,  if  the  shear  be  known,  and  the 
angle  included  between  the  vertical  and 
a  diagonal  be  6,  we  have  only  to  multi- 


13 

ply  the  shear  by  sec.  6  to  find  the  strain 
in  the  diagonal.  Again,  let  the  section 
be  moved  to  the  left  so  as  to  pass  through 
the  point  e,  and  let  that  point  be  taken  as 
the  center  of  moments.  Then  the  mo 
ment  of  resistance  will  be  the  moment 
of  the  chord  strain  E  F.  Hence,  if  that 
moment  be  known,  we  have  only  to  di 
vide  it  by  the  depth  of  the  truss  to  get 
the  strain  in  E  F. 

The  internal  shear  and  moment  at 
any  section  are  easily  found  from  the 
fundamental  conditions  of  statical  equili 
brium.  The  shear  being  an  internal 
vertical  force  is  the  resultant  of 
the  exterior  vertical  forces  on  either 
side  of  the  section.  The  exterior  forces 
on  the  left  of  the  section,  for  instance, 
have  for  a  resultant  their  algebraic  sum  ; 
considering  the  upward  forces  as  posi 
tive,  and  the  downward  ones  as  negative 
we  have  from  Fig.  1,  their  sum 


—  wx 


as  the  expression  for  the   shear  in  the 
section  x.     To  get  the  internal  moment 


14 

for  the  same  section,  we  have  only  to 
consider  in  like  manner  that  it  is  equal 
to  the  sum  of  the  moments  of  all  the  ex 
terior  forces  on  either  side  of  the  section, 
for  if  otherwise,  there  would  be  a  ten 
dency  to  rotation.  The  moment  of  the 
force  Ri  with  reference  to  x  is  R: 
(I,  +  Za  +  a)  ,  of  R2  is  R2  (I,  +  x)  ,  of  the  load 
t0il9  iatoi,  (i^  +  ^+aJ),  etc.  Thus  from 
a  mere  inspection  of  Fig.  1  we  write  the 
value  of  the  moment  M,  regarding  those 
moments  as  positive  which  cause  a  ten 
sile  strain  in  the  upper  fiber  at  x,  and 
those  as  negative  which  cause  a  compres- 
sive  one.  The  expression  is 


+  x)  +  w  I*  (i  k  +  x)  —  1^3  x  +  wx\  x 
Now  in  these  expressions  for  the  in 
ternal  shear  S  and  the  moment  M  at  any 
point  x,  the  lengths  £,  £2,  x  are  given  by 
the  conditions  of  the  case  in  hand,  and 
the  same  is  true  of  the  load  per  linear 
unit  w.  Hence  the  shear  and  moment, 
and  consequently  the  internal  strains  are 
easily  obtained  as  soon  as  the  reactions 


15 

of  the  supports  are  known.  We  shall 
hereafter  give  methods  by  which  the  re 
actions  may  be  readily  determined. 

By  the  same  reasoning  if  we  pass  a 
section  in  the  span  Z2,  (Fig.  1)  at  a  point 
whose  distance  from  the  support  2  is  x, 
the  shear  S  and  the  moment  M  for  that 
section  will  be 


W  X* 


^^—  w  2—  w  x 

Mc=  -Rx  (Z,  +  x)  +  w  1  1  (\  1,  +  x) 


In  a  simple  girder  whose  length  is 
each  of  whose  reactions  is  R  or  ^  w  I  the 
shear  and  moment  for  any  section  x  will 
be 

I—  x) 


WX*       W 


2 


When  the  number  of  spans  is  large, 
the  expressions  for  the  shear  and  moment 
as  above  deduced  become  long  and  in 
volve  much  arithmetical  computation. 
We  are,  however,  fortunately  able  to 


16 


place  them  under  a  much  simpler  form. 
First  they  may  be  written  thus 


—  wx 
M=  [-K,   (£  + 


wl^  +  R3)  is  the 
shear  in  the  span  13  at  a  point  infinitely 
near  to  the  support  3  ;  let  this  be  called 
S3.  Also  the  quantity  enclosed  in  [  ]  in 
the  second  equation  is  the  moment  of 
the  exterior  forces  with  reference  to  the 
point  3  ;  let  this  be  called  M3.  Then  the 
equations  become 


Therefore  the  internal  shear  and  mo 
ment  at  any  section  can  immediately  be 
found)  without  the  necessity  of  determin 
ing  the  reactions,  provided  ice  know  the 
shear  and  the  moment  for  the  preceding 
support.  This  method,  due  to  Clapeyron, 
of  using  the  moment  at  the  supports  in 
stead  of  the  reactions  greatly  simplifies 


17 

the  numerical  computations  of  a  continu 
ous  truss.  We  designate  the  moment  at 
3  by  M3,  the  reaction  being  R3,  or  the 
sum  of  the  shear  S3,  in  the  span  1 3  and 
of  the  shear  S'2  in  the  span  1 2  both  infi 
nitely  near  to  the  support  3.  In  like 
manner  the  moments  at  the  supports  2 
and  4,  will  be  designated  by  M2  and  M  , 
the  shears  just  to  the  right  of  those 
points  by  S2  and  S4,  and  those  to  the  left 
by  S'j,  and  S'3.  In  general  for  any  sup 
port  whose  index  is  n,  wre  have  (Fig.  3) 

Fig.  3, 

<• 7 * 7 *-  7.  _  _ 

-u-i  n     I  ^      n+i 

n-\ ^ n-t-i 

^^^^^"^^^ji— -X 


- 


on  the  left,  the  span  £n_i,  on  the  right  the 
span  /n  ;  the  shear  infinitely  near  to  n 
on  the  left  is  S/n_i,  on  the  right  Sn  ;  the 
sum  of  S'n_i,  and  Sn  makes  the  reaction 
Rn  ;  and  the  moment  over  the  support 
is  MG.  If  the  load  be  uniform  and 
equal  to  w  per  linear  unit,  the  internal 


18 

• 

shear  S  and  moment  M  for  any  section  x 
are  given  by 

S  =  Sn  —wx 

M=Mn  —  Sn  x  + 1  w  x* 

To  find  then  the  internal  strains  in 
the  diagonals  of  a  continuous  truss  due 
to  dead  load  only,  we  have  to  pass  a 
section  cutting  each  diagonal  and  find 
the  value  of  S,  this  is  the  shear  which 
the  strain  in  the  diagonal  must  resist 

o 

and  multiplied  by  sec.  8  (8  being  the  in 
clination  of  the  diagonal  to  the  vertical,) 
it  gives  the  required  strain.  To  find  the 
strains  in  the  upper  chord  we  have  to 
take  the  lower  chord  apices  as  centers 
of  moments  and  compute  the  values  of 
M  ;  these  divided  by  the  depth  of  the 
truss  give  the  strains,  which  are  tensile 
if  M  is  positive,  compressive  if  M  is  neg 
ative.  To  find  the  lower  chord  strains, 
we  choose  the  upper  apices  from  which 
to  measure  the  values  of  a?,  and  divide 
the  resulting  values  of  M  by  the  depth 
of  the  truss  ;  if  M  is  positive  these  give 
compressive  strains  ;  if  negative,  tensile 
ones. 


Everything  is  thus  known,  except  the 
shears  and  moments  at  the  supports,  and 
for  these  formulae  and  methods  will  be 
presented  in  Chapter  II,  by  which  they 
may  be  found  for  all  cases.  We  give 
here,  however,  two  tables  from  which 
they  may  be  found  for  the  common 
case  when  all  the  spans  are  equal,  and 
which,  by  a  simple  law,  may  be  extend 
ed  to  include  any  number  of  such  spans. 

As  before,  let  w  be  the  unifornily  dis 
tributed  load  per  linear  unit;  let  I  be  the 
length  of  each  span,  then  will  w  I  be  the 
weight  of  one  span.  The  shear  at  any 
support  is  a  fractional  part  of  w  I,  or 

s  A  w  I 


A  being  a  fraction  given  in   the  follow 
ing  triangle  : 


20 


21 


Each  of  the  squares  composing  this 
triangle  represents  one  of  the  supports, 
and  its  left  hand  division  gives  the  left 
hand  shear  S'n_i,  and  the  right  hand  one 
the  shear  Sn  (Fig.  3).  Thus,  in  the  gir 
der  of  three  spans,  the  triangle  shows 
that  the  first  support,  beginning  at  the 
left,  has  on  the  left  no  shear,  and  on  the 
right  YQ  10  /,  that  the  second  support  has 
on  the  left  a  shear  of  ^0  w  /,  and  on  the 
right  one  of  -f0  w  I.  The  sum  of  the  two 
shears  for  any  supports  is  of  course  its 
reaction.  For  example,  a  girder  of  six 
equal  spans  has  at  its  middle  support  a 
reaction  of  I8t  w  I. 

The  moment  at  any  support  will  be  a 
fractional  part  of  ivl^or 

7. 

Moment  =  B  w  l^ 

B  being  a  fraction  given  in  the  following 
triangle  ;  in  which  like  the  preceding 
one  the  spaces  indicate  the  supports  of 
the  girder.  Thus,  the  fourth  horizontal 
line  refers  to  a  girder  of  four  spans,  the 
moments  at  the  first  and  last  supports 
being  0,  at  the  second  and  fourth  -238-  w  I* 
and  at  the  middle  one  w  I2. 


22 


23 


The  triangles  can  be  extended  to  any 
required  length  by  the  application  of  the 
following  law  which  obtains  in  &\\  oblique 
columns.  Any  fraction  belonging  to  an 
even  number  of  spans,  may  be  obtained 
by  multiplying  both  numerator  and  de 
nominator  of  the  preceding  fraction  by 
two  and  adding  the  numerator  and  de 
nominator  of  the  fraction  preceding  that. 

33 
Thus  for   eight  spans,  the  fraction 


2X11  +  11 

is  equal  to  -  —  -  or  to 

2X142  +  104 


<388 


,. 
2X142  +  104'  accordmS  as  we  use   one 

oblique  column  or  the  other.  For  an  odd 
number  of  spans,  any  fraction  is  found 
by  adding  the  two  preceding  fractions, 
numerator  to  numerator  and  denomina 
tor  to  denominator.  Thus  for  seven 


spans, 


12          8+4  12          9  +  3 

or 


142     104  +  38        142     104  +  38 

These  tables*  furnish  the  data  for  solv- 

*  These  triangles  were  first  given  by  the  author  in  the 
Journal  of  the  Franklin  Institute  for  March,  1875.  A  de 
monstration  of  the  laws  governing  them  may  be  seen  in 
the  same  Journal  for  April,  1875. 


24 


ing  all  questions  concerning  continuous 
girders  whose  supports  are  on  the  same 
level,  whose  spans  are  all  equal,  and 
which  are  loaded  uniformly  throughout 
their  entire  length.  The  reader  should 
first  acquire  facility  in  the  use  of  the 
tables.  We  give,  therefore,  a  few  ex 
amples  for  practice  : 

1.  In  a  girder  of  six  spans,  what  is  the 
reaction  at  the  second  support  ? 

118 

Ans.  K.  =  -—  -  w  I. 
104 

2.  In  one  of  eight  spans,  what  is  the 
reaction  at  the  middle  support? 

386 

Ans.K=-wl 

3.  In  one  of  ten  spans,  what  is  the  mo 
ment  over  the  fourth  support  from  the 
left? 


4.  In  one  of  seven  spans,  what  are  the 
shears  S2  and  S'?     (see  Fig.  3.) 


25 


5.  In  one  of  six  spans,  what  is  the  mo 
ment  M   and  the  shear  S6  ? 


Ans.  M= 


Having  thus  found  from  the  triangles, 
.the  moment  Mn  and  the  shear  Sn  for  the 
nth  support,  the  shear  S  and  the  moment 
M  for  any  section  in  the  nih  span  are 
readily  found  from  the  formulae 

O  - —  On  —  W  X 

Tf  C       ™     I      1     nn  rf 

—  -LT_Ln  —  c?n  th  ~r  2^  w  **j 

which  we  have  demonstrated  above,  and 

Fig.  3. 

n-i       n         n+i 


in  which  x  is  the  distance  from  the  sup 
port  n  to  the  assumed  section.  If  in 
these  x  be  made  equal  to  I,  they  will,  of 
course,  give  the  shear  S'n  at  the  left  of 
the  w  +  lth  support,  and  the  moment 
Mn+i  over  that  support.  We  will  illus 
trate  their  use  by  a  few  examples  : 


26 


6.  In  a   continuous   girder   of    three 
spans,  what  is  the  shear  and  the  moment 

at  the  center  of  the  middle  span  ? 

5 

We  have  from  the  table  Sa=---  wl  and 
1  10 

M,=-«r. 

Hence 

S==  —  w  I—  iv  x 

1  5 

M=—  ivF——-w  l 

and  placing  x  equal  to  £  /,  we  have 

S=0     M=-i«rP 

40 

7.  In   a  girder  of  six  spans,  find   the 
shear  and  moment  at  the  center  of  the 
second  span  ? 

Ans.  S=—Awl     M  =  --  -  w  ? 
104  208 

8.  In  one  of  four   spans,  what  is  the 
shear  and  moment  in  the  third  span  for 
x-=.\  I  and  x=%l? 

Ans.     S=  —  wl     and  S=  --  wl 
14  14 


27 

In  computing  a  framed  truss  we  need 
to  find  the  value  of  S  for  a  section  cut 
ting  every  diagonal  and  that  of  M  for 
one  passing  through  each  panel  apex, 
from  these  we  readily  derive  the  strains 
in  the  webbing  and  chords  by  the  rules 
explained  above.  A  single  example  will 
render  the  whole  process  clear.  (We 
here  treat  of  the  dead  load  only  ;  com 
putations  involving  the  live  or  rolling 
load  will  be  presented  hereafter.) 

Let  the  truss  represented  in  Fig.  2 
consist  of  seven  continuous  spans,  each 
sixty  feet  in  length.  Let  the  uniformly 
distributed  load  per  linear  foot  be  two 
hundred  pounds,  one  half  of  which  rests 
upon  the  upper  chord  and  the  other  half 
upon  the  lower.  The  lower  chord  is  di 
vided  into  six  bays,  each  of  ten  feet,  and 
is  connected  with  the  upper  one  by  a 
Warren  system  of  diagonals.  The  depth 
of  the  truss  is  seven  feet.  Let  it  be  re 
quired  to  compute  the  strains  in  all  the 
pieces  of  the  third  span,  due  to  this  dead 
load. 

We  first  take  from  the  triangles  for  a 


28 


girder  of  seven  spans.  Sq  =  —  wl     and 


M3=-—  wl*.     We   have  for   our   case, 

w=200  Ibs.  and  1=60  feet.     Hence 
S3  —  5916  Ibs.  andM3=55775  Ibs.  ft. 

Inserting  these  and  the  value  of  w  in  the 
above  general  formulae,  we  have 

S  =  5916  —  wx 

M  =  55775  —  5916  ^  +  100  x* 

as  the  value  of  the  shear  and  moment  for 
anv  section  x. 

v 

Now,  since  this  is  a  framed  truss,  and 
the  several  pieces  are  to  be   subjected 


only  to  longitudinal  strains,  the  load 
should  not  be  strictly  uniformly  distrib 
uted  but  concentrated  on  the  upper 
chord  at  the  panel  points  B,  C,  etc.  and 
on  the  lower  chord  at  a,  b,  c,  etc.  Allow 
ing  that  each  of  these  points  receives  an 


29 

equal  weight  and  that  a  and  g,  count  as 
but  one  point,  we  have  at  a  500  Ibs.,  at 
g  500  Ibs.  and  at  each  of  the  others  1000 
Ibs.  In  finding  the  shear  for  the  diag 
onal  a  B,  we  pass  the  section  anywhere 
between  a  and  B  and  takewx  as  500, 
for  B  b  10  x  is  1500,  for  b  C  2500  and  so 
on  ;  these  subtracted  from  S3  give  the 
required  shears.  This  is  in  fact  nothing 
but  taking  the  algebraic  sum  of  all  the 
exterior  forces  between  the  left  hand  of 
the  truss  and  the  diagonal  under  consid 
eration,  for  S3  is  the  sum  of  those  forces 
from  the  left  end  to  the  beginning  of  the 
span.  For  the  diagonal  Fe  we  have,  for 
example, 

S  =  5916  -  -  8500  =  -  -  2584  Ibs. 

Thus  by  successive  subtraction  we 
find  the  shears  for  all  diagonals.  Multi 
plying  them  by  the  secant  of  the  angle 
between  a  diagonal  and  the  vertical  or 


30 


and  we  have  the  required  strains.  To 
determine  their  character  we  have  simply 
to  consider  that  a  positive  shear  causes 

(        tensile       ) 

a   -{  •      I    strain  in  a  diagonal 

(  compressive  j 

which  slopes  \    uPward    \     toward    the 
(  downward  ) 

left  hand  support,  while  a  negative  shear 
produces  the  reverse.  In  the  following 
table,  the  results  thus  determined  are 
recapitulated.  The  first  column  shows 
the  name  of  the  diagonal  corresponding 
to  Fig.  2,  the  second  gives  the  shears, 
the  third  gives  the  slopes,  +  indicating 
an  upward  inclination  toward  the  left, 
and  -  a  downward  one,  and  the  last 
column  gives  the  strains,  -f  indicating 
tension,  and  —  compression.  In  forming 
the  last  column  from  the  two  preceding 
ones,  it  will  be  noticed  that  the  rule  of 
signs  is  observed  : 


31 


DIAGONALS.     (See  Fig.  2.) 


Piece. 

Shear. 

Slope.    Strain. 

Cb 

J 

-5416 
-3416 

+ 

-66561bs. 

+5427 
-4198 

Cc 
DC 

h2416 
-1416 

± 

-U2969 
-1740 

Ed 

+  416 
-  584 

-1584 

- 

+  511 
+  718 
-1947 

Ye 

-2584 
-3584 

^ 

+3176 
-4405 

S{ 

-4584 

+5634 
-6863 

We  will  now  pass  to  the  computation 
of  the  chord  strains.  In  the  above  ex 
pression  for  M  the  quantity  ^  10  cc2  or 
100  x*  is  the  moment  of  the  load  between 
the  point  3  and  the  assumed  section  and 
its  value  is  the  same  whether  the  load  be 
considered  as  uniformly  distributed  or 
concentrated  at  the  apices  as  above. 
Hence  to  find  the  moments  for  the  upper 
chord  we  have  in  the  expression 

M  —  55775  --  5916^+lOOar2 
simply  to  give  to  x  the  successive  values 


32 

0,  10,  20,  etc.,  since  for  the  bays  A  B, 
B  C,  C  D,  etc.,  the  opposite  vertices  #,  b,  c 
etc.,  must  be  taken  as  centers  of  moments. 
Thus  if  the  bay  C  D  be  cut  rotation  will 
at  once  begin  around  the  point  c  ;  we 
take  then  x  =20  and  find  for  the  mo 
ment  of  the  strain  in  C  D, 

M=  -  -  22545  Ibs.  ft. 

and  dividing  this  by  its  lever  arm  or 
seven  feet  we  have  3221  Ibs  for  the 
strain.  The  character  of  the  strain  is 
found  by  recollecting  that  a  positive  mo- 

(       tensile       } 

ment  causes  a    \  .     >    strain  in 

(  compress^ve  \ 

the          **       [    chord,    while   a  negative 
(  lower  } 

moment  produces  the  reverse.  If  we 
designate  then  tension  by  -f  and  com 
pression  by  — ,  the  signs  of  the  strains 
in  the  upper  chord  will  be  the  same  as 
those  of  the  moments.  In  this  way  it  is 
easy  to  compute  the  following  results  : 


33 


UPPER  CHORD.     (See  Fig.  2). 


Bay. 

Moment. 

Strain. 

AB 
BC 
CD 
DE 
EF 
FG 
GH 

+55775  Ibs.  ft. 
+  6615 
-22545 
-31705 
-20865 
+  9985 
+60845 

+7968  Ibs. 
+  945 
-3221 
-4529 
-2981 
+1426 
+8692 

For  the  lower  chord  the  calculation  is 
very  similar.  The  centres  of  moments 
are  taken  at  the  points  B,  C,  etc.,  the 
successive  values  of  x  are  5,  15,  25,  etc., 
the  strains  are  numerically  one-seventh 
of  the  moments,  and  their  signs  are  op 
posite  to  that  of  the  moments.  Thus  for 
the  bay  e/,a?=45,  M=  -  7945  Ibs.  ft. 
and  the  strain  in  ef  is  1135  Ibs.  tension. 
The  results  are  in  the  following  table  : 


34 


LOWER  CHORD.    (See  Fig.  2.) 


Bay. 

Moment. 

Strain. 

ab 

+28695  Ibs.  ft. 

-4099  Ibs. 

be 

-10465 

-1495 

cd 

-29625 

-4218 

de 

-28785 

_ 

-4112 

•f 

-  7945 

-1135 

fg 

+32895 

-4699 

and  the  strain  sheet  for  the  span  is  now 
complete. 

In  the  same  way  the  strains  for  each  of 
the  other  spans  may  be  readily  found. 
From  the  symmetry  of  the  truss  it  is 
evident  that  the  fifth  span  will  be  exact 
ly  the  same  as  the  third,  the  sixth  the 
same  as  the  second,  and  the  seventh  the 
same  as  the  first.  For  the  fourth  span 
the  value  of  S  and  M  for  any  section  x 

are 

w  x 


M=60848  —  6000&  +  100  x3 
and  the  strains  will  be  the  same  on  each 
side  of  its  center.     For   the  first   span 
Sj  is  the  same  as  the  reaction  R1  the  mo 
ment  M,  is  zero  and  we  have 


35 


8  =  4732  —  wx 
M=  —4732  x+100 

It  will  be  seen  then  that  the  compu 
tation  of  the  strains  in  a  continuous 
girder  is  exactly  the  same  as  in  a  simple 
one,  except  only  in  the  preliminary  de 
termination  of  the  shears  and  moments 
at  the  supports.  In  a  simple  girder  the 
end  shears  are  the  same  as  the  reactions 
which  are  known  from  the  law  of  the 
lever,  and  the  moments  at  the  supports 
are  zero.  In  a  continuous  one  these 
quantities  must  be  determined  by  formu 
lae,  or,  for  the  case  of  equal  spans  uni 
formly  loaded,*  taken  from  the  triangles 
which  we  have  given  above.  They  may 
also  be  found  by  a  graphical  process. 

If  the  truss  above  discussed  were  built 
with  seven  simple  girders,  the  strains  in 
each  would  be  the  same.  It  may  prove 
interesting  then  to  compare  the  results 
above  found  with  those  for  a  simple  gir- 

*  Other  convenient  tables  for  concentrated  loads  and 
for  uniform  loads  over  one  span  only  are  given  in  the  ar 
ticle  above  referred  to  in  the  Journal  of  the  Franklin  In. 
stitute. 


36 


der.  The  mode  of  computation  is  essen 
tially  the  same  ;  the  end  shears  are  each 
one  half  the  total  load,  the  pieces  A  B 
and  G  PI  in  Fig.  2  disappear,  and  for  any 
section  x  we  have 

8=6000—  wx 


Considering  as  before  that  the  load  is 
concentrated  at  the  panel  points  we  have 
500  Ibs.  at  a  and  y  and  1000  Ibs.  at  B,  C, 
D,  £,  c,  d,  etc.,  respectively.  We  then 
find  the  shears  and  moments  and  from 
them  deduce  the  strains  as  above  de 
scribed.  The  results  are  given  below 
compared  with  those  for  the  third  span 
of  the  continuous  truss. 

(See  Table  on  following  pages.} 

Adding  these  strains  regardless  of 
sign  we  find  the  two  sums  to  be  the 
same.  It  can  be  easily  demonstrated 
that  for  the  dead  load  such  should  be 


DIAGONALS.     (See  Fig.  2.) 


Piece. 

Continuous 
truss. 

Simple  truss. 

Ea 

-6656  Ibs. 

—6760  Ibs. 

Cb 

+5427 
-4198 

+5531 
—4301 

Cc 
DC 

+2969 
—1740 

+3072 
—1844 

Dd 

Erf 

+  511 
+  718 
-1947 

+  614 
+  614 
—1844 

¥e 

+3176 
-4405 

+3072 
—4301 

G£ 

+5634 
—6863 

+5531 
—6760 

Sams. .    .  44244  Ibs. 


44244  Ibs. 


the  case.  As  far  as  the  diagonals  are 
concerned,  the  two  structures  require  an 
equal  amount  of  material. 

UPPER  CHORD.     (See  Fig.  2.) 


Bay. 

Continuous 
truss. 

Simple  truss. 

AB 
BC 
CD 
DE 
EF 
FG 
GH 

+7968  Ibs, 
+  945 
—3221 
—4529 

—2981 
'  +1426 
+8692 

—  7143  Ibs. 
—11429 

—12857 
—11429 
—  7143 

Sums.  .    .  29762  Ibs. 

50001  Ibs. 

38 


Adding  the  strains  in  the  upper  chord 
we  observe  that  the  sum  for  the  simple 
truss  is  about  1.7  times  that  of  the  other. 
If  the  amount  of  material  is  to  be  pro 
portional  to  the  strain,  a  considerable 
saving  will  here  be  expected. 

LOWER  CHORD.     (See  Fig.  2.) 


Bay. 

Continuous 
truss. 

Simple  truss. 

ab 

-4099  Ibs. 

—  3929  Ibs. 

be 

-1495 

—  9643 

cd 

-4218 

—12500 

de 

-4112 

—12500 

ef 

-1135 

—  9643 

fff 

-4699 

—  3929 

Sum....  19758  Ibs. 

52144  Ibs. 

The  lower  chord  in  the  simple  truss 
would  then  be  subjected  to  about  2.6 
times  as  much  strain  as  in  the  continuous 


one. 


If  we  suppose  that  the  same  working 
strength  may  be  allowed  for  compression 
as  for  tension,  we  may  obtain  an  estimate 
of  the  saving  in  material  by  employing  a 


39 


continuous  truss  instead  of  a  simple  one. 
The  amount  of  material  will  be  propor 
tional  to  the  strain  and  to  the  length  of 
the  piece  strained.  Regarding  the  bays  of 
the  chord  as  unity,  the  diagonals  will  be 
represented  in  length  by  0.86.  The  pro 
portionate  amounts  of  iron  will  then  be 
found  by  multiplying  the  above  sums  by 
unity  for  the  chords  and  by  0.86  for  the 
diagonals.  Thus  we  have  a 

COMPARISON. 


Continuous 
truss. 

Simple  truss. 

Diagonals  .... 
Upper  Chord.  . 
Lower  Chord.  . 

38050 
29762 
19758 

38050 
50001 
52144 

Total  

87570 

140195 

from  which  we  see  that  the  amounts  of 
material  in  the  two  cases  are  in  the  ratio 
of  the  numbers  87570  and  140195  or  as 
1  to  1.6.  For  this  particular  span  then 
a  saving  in  material  of  thirty-seven  and 
a  half  per  cent,  is  effected  by  using  a  con 
tinuous  truss  instead  of  a  common  one. 


40 

It  is  capable  of  demonstration  that  for 
girders  subjected  only  to  dead  load,  the 
total  amount  of  strain  in  the  webbing 
will  be  the  same  for  simple  as  for  con 
tinuous  trusses,  and  also  that  under  the 
most  favorable  circumstances,  the  total 
strain  in  the  chords  of  the  first  is  to  that 
in  the  chords  of  the  second  as  /\/2^  is  to 
2  or  nearly  as  2.6  to  1. 

In  studying  the  theory  of  girders  many 
interesting  questions  arise  which  are  of 
little  importance  in  practice.  One  of 
these  is  the  determination  of  the  inflec 
tion  points.  At  these  points  the  curva 
ture  of  the  beam  changes,  the  strain 
passes  from  tension  to  compression  and 
the  moment  is  zero.  At  any  point  in  the 
nth  span  the  moment  is 

M  =  Mn  -  -  Sn  x  +-J-  w  x* 

Making  in  thisM  equal  to  zero  and  solv 
ing  the  equation  with  reference  to  x  we 
find 


;2  "       w 


41 


For  a  girder  of  equal  spans  and  uniform 
ly  loaded  we  may  hence  write  for  the  two 
inflection  points, 


in  which  A  and  B  are  to  be  taken  from 
the  above  triangles,  A  always  being  taken 
for  the  right  hand  side  of  the  support 
under  consideration,  for  example,  in  a 
girder  of  eight  spans  the  inflection  points 
for  the  fourth  span  are  at  the  points 

195 


V388/        388 
or  for  33=0.22  I  and  ^=0.79  L 

The  point  of  maximum  moment,  or 
the  point  near  the  center  of  the  beam, 
where  the  chord  strain  is  the  greatest  is 
more  important  and  readily  determined 
from  the  above  general  value  for  M. 
Differentiating  it  with  reference  to  x  we 
have 


=  —  Sn  +  W  23=0 

a  x 
that  is,  the  maximum  moment  obtains  at 


42 


the  point  where  the  shear   is  zero.     Its 

S 
value  is  found   by  replacing  x   by 

which  gives 

ft2 
Max.  M=Mn  — 


2  w 

as  the  greatest  negative  moment. 

The  following  examples  will  enable 
the  reader  to  test  his  knowledge  of  the 
preceding  principles  : 

9.  In  a  girder  of  two  spans  uniformly 
loaded  what  are  the  maximum  positive 
and  negative  moments  ? 

Ans.  0.125  wT  and— 0.071  wl\ 

10.  In  one  of  three  spans  what  is  the 
maximum  negative  moment  in  the  mid 
dle  span  ? 

Ans.   —0.025  wF. 

11.  In  one  of  eight  spans  where  are 
the  inflection  points  in  the  fifth  span  ? 

Ans.  x=  0.2lZandcc=( 


12.  A  continuous  girder  of  three  spans, 
each  equal  to  fifty  feet,  is  divided  into 


43 

five  panels  on  the  lower  chord,  and  has 
bracing  similar  to  that  shown  in  Fig.  2. 
Supposing  a  load  of  five  tons  applied  at 
each  of  the  lower  panel  points,  what  are 
the  strains  in  each  of  the  pieces  of  the 
middle  span  ?  the  height  of  the  truss 
being  six  feet. 

Ans.  In  a  &,  --11.7  tons;  in  b  c,  +1.7; 
in  B&,  +13  ;  etc. 

In  this  chapter  we  have  treated  of  the 
continuous  girder  when  affected  only  by 
dead  load  or  its  own  weight.  In  the  fol 
lowing  chapters  we  shall  take  up  the  ac 
tion  of  the  live  or  variable  load. 


45 


CHAPTER    II. 

A  continuous  girder,  loaded  in  any 
manner,  is  held  in  equilibrium  by  the  up 
ward  pressures  or  reactions  of  the  sup 
ports,  and,  as  we  have  seen,  these  reac 
tions  are  alone  sufficient  for  the  complete 
determination  of  the  strains  in  every 
part  of  the  girder.  But  if,  regarding  the 
question  as  one  of  pure  statics,  we  con 
sider  the  beam  as  rigid,  we  find  it  impos 
sible  to  determine  the  reactions  when  the 
number  of  supports  is  greater  than  two. 
This  does  not  arise  from  the  fact  that  in 
an  actual  case  the  question  is  indeterm 
inate,  but  simply  because  in  considering 
the  girder  as  rigid  we  have  restricted  the 
data  to  the  mere  weight,  neglecting  en 
tirely  the  physical  properties  of  the  ma 
terial.  By  taking  into  account  the  elas- 


46 

ticity  of  the  girder,  the  problem  becomes 
determinate  ;  we  find  the  reactions,  or 
what  is  equivalent,  the  shears  and  mo 
ments  at  the  supports,  and  from  these 
the  investigation  of  the  internal  forces 
or  strains  is  easy. 

THE    ELASTIC    LINE. 

When  a  girder  is  acted  upon  by  verti 
cal  forces,  a  change  of  shape  arises, 
which  causes  the  originally  parallel  fibers, 
to  be  on  one  side  lengthened,  and  on  the 
other  shortened.  Between  the  lengthen 
ed  and  shortened  fibers,  there  is  a  plane 
which  undergoes  no  change  of  length  ; 
the  central  line  of  this  plane  is  called  the 
neutral  axis  or  the  elastic  line.  Thus,  in 
the  bent  beam  represented  in  Fig.  4,  m  o 
is  the  neutral  axis,  the  fibers  above  it  be 
ing  shortened  "or  compressed,  and  those 
below  it  lengthened  or  tensioned. 

We  derive  the  equation  of  the  elastic 
line  upon  three  hypotheses  :  1st,  that  all 
planes  perpendicular  to  the  neutral  axis 
before  the  bending  or  flexure,  preserve 


4 


- 


during  the  bending  their  perpendiculari 
ty  and  their  form  as  planes  ;  2d,  that  the 
change  of  length  of  a  body  subjected  to 
a  force  is,  within  certain  limits  called  the 
elastic  limits,  proportional  to  the  intensi 
ty  of  the  force  ;  and  3rf,  that  the  change 
of  shape  is  so  little  that  the  length  of  the 
neutral  axis  is  sensibly  the  same  as  its 
horizontal  projection. 

In  Fig.  4  we  have  a  longitudinal  sec 
tion  of  a  portion  of  a  bent  beam ;  the  two 
planes  a  b  and  d  6,  originally  parallel,  re 
maining  perpendicular  to  the  neutral 
axis  m  o,  and  intersecting  in  c  the  center 
of  curvature.  Hence,  drawing  f  g  paral 
lel  to  ab  through  o,  the  lines /V7,  ge,  etc., 
denote  the  elongations  and  compressions 
of  the  respective  fibers,  and  we  see  from 
the  figure  that 

od:odr  \\df:  d' f 

or  the  change  of  length  in  the  fibers  is 
proportional  to  their  distances  from  the 
neutral  axis.  This  is  the  consequence  of 
the  first  hypothesis. 

Designating  by  H  and  H'  the  force 


48 

acting  in  the  fibers  df  and  d'  /",  the  sec 
ond  hypothesis  says  that 

H:H':;<7/:  d'  f 
FIG.  4. 


Hence,  by  combining  these  two  propor 
tions, 

H  :  H'  ;  I  o  d  :  o  dr 
or,  the  horizontal  forces  are  directly  pro- 


portional  to  their  distances  from  the 
neutral  axis.  Therefore,  if  we  denote 
the  distance  of  any  fiber  from  the  axis 
by  2  the  strain  upon  it  by  H',  the  dis 
tance  of  the  remotest  fiber  by  e  and  the 
strain  upon  it  by  H,  we  have 

H'  :  H  ;  ;  z  :  e     or  H'=  — 

e 

We  have  thus  far  considered  the  cross- 
section  of  the  fibers  as  unity.  If  the 
actual  section  be  a,  the  force  in  each  is 

evidently  -      -  .     Each   of   these  forces, 

as  for  instance  H'  in  the  figure,  tends  to 
turn  the  beam  around  the  point  o  with  a 
lever  arm  o  d'  or  z,  and  its  moment  or  the 
measure  of  that  tendency  to  rotation  is 

the  product  of  the  force          *  by  the  dis- 

TT  y  8 

tance  z,  or  -  .     The  sum  of  all  these 


moments  is 


M=- 

e 


50 


or,  since  2  a  £  is  the  moment  of  inertia 
of  the  section  a  &,  we  have 


6 

as  the  expression  for  the  sum  of  the  mo 
ments  of  the  internal  forces,  H  being  the 
strain  in  the  remotest  fiber,  e  its  distance 
from  the  neutral  axis,  and  I  the  moment 
of  inertia  of  the  cross-section. 

The  line  df  denotes  the  change  of 
length  of  the  fiber  ad  due  to  the  force 
H.  Hence  if  E  be  the  coefficient  of 
elasticity,* 

ad:  df:  ;  E:  H 

Designating  the  radius  c  o  by  r  we  have 
from  the  similar  figures  o  df  and  cad 
(since  mo=ad), 

a  d  :  df.  \r  :  e 

*  The  Coefficient  of  Elasticity  is  the  ratio  of  the  force 
of  displacement  to  the  amount  of  displacement  taken 
upon  a  cube  whose  edge  is  unity  ;  Hence  for  the  above 

J          4* 

case  E=H-$  —  -  .    The  term  modulus  of  elasticity  pro- 

wtt 

perly  relates  to  the  impact  of  bodies,  and  is  a  measure  of 
elasticity  in  the  common  sense  of  the  word,  unity  indi 
cating  perfect  elasticity  or  restitution  of  form.  These 
terms  are  often  confounded  by  writers. 


51 

Combining  these  proportions  we  find 

H_E 

e      r 

and  hence,  for  the  value  of  the  internal 
moment,  we  have 


The  radius  of  curvature  of  any  plane 
curve,  whose  length  is  u,  and  co-ordinates 
x  and  y  is* 


r= 


d  x  d*  y 

And  as  by  our  third  hypothesis  we  may 
place  du=dx,  this  becomes 

r_dx* 

/*y     ^  i 

\At       */ 

which,  inserted  in  the  above  value  of  M, 
gives 


(1) 


d*  y       M 


dx*      El 
as  the  differential  equation  of  the  elastic 

*  See  any  work  on  the  Differential  Calculus. 


52 

curve,  applicable  to  all  bodies  subject  to 
flexure,  which  fulfill  the  condition  im 
posed  by  the  third  hypothesis.  The  co 
efficient  of  elasticity  and  the  moment  of 
inertia  may  be  different  in  every  section. 

CONDITIONS    OF    EQUILIBRIUM. 

Let  us  consider  the  rthv  span  of  a  con 
tinuous  girder  whose  length  is  lTy  and 
let  a  single  concentrated  load  Pr  be 
placed  on  this  span  at  a  distance  Mr  from 
the  left  hand  support  r.  This  load,  the 
loads  on  the  other  spans,  and  the  weight 
of  the  girder  itself,  are  held  in  equili 
brium  by  the  vertical  reactions  Rr— i, 
RTj  etc.,  of  the  several  supports.  (See 
Fig.  6.) 

Let  us  pass  a  section  between  the  load 
Pr  and  the  support  r-j-1  at  a  distance  x 
from  the  rth  support.  As  shown  in  the 
last  chapter,  all  the  internal  forces  in  this 
section  are  represented  by  a  shear  S  and 
a  moment  M.  The  shear  S  is  equal  to 
the  algebraic'  sum  of  all  the  external 
forces  upon  the  left  hand  side  of  the  sec- 


53 

tion,  and  the  moment  M  is  equal  to  the 
sum  of  the  moments  of  those  forces  with 
respect  to  the  section  as  a  center.  Hence, 
regarding  upward  forces  as  positive,  and 
a  moment  as  positive  when  it  tends  to 
cause  tension  in  the  upper  fiber  of  the 
section,  wre  have 

(2)  S  =  S,  -  Pr 


in  which  Sr  is  the  shear  at  the  right  of 
the  rth  support,  and  Mr  the  moment  at 
that  support.  In  like  manner  for  a  sec 
tion  between  r  and  Pr>  we  have 

S=Sr 

M=Mr  -  -  Sr  x 

The  internal  forces  at  any  section  can 
then  be  found  as  soon  as  the  shear  and 
moment  at  the  preceding  support  are 
known. 

If,  in  the  above  expression,  we  make  x 
equal  to  /r,  M  becomes  Mr  _j_  i,  and  we  de 
duce 


54 

The  shear  and  moment  at  any  section 
can  then  be  determined  as  soon  as  Mr  and 
Mr  4_  i,  the  moments  at  the  preceding  and 
following  supports,  are  known. 

These  conditions  of  equilibrium  are 
entirely  independent  of  variations  in  the 
dimensions  or  material  of  the  beam,  or 
in  the  relative  heights  of  the  supports  of 
tlie  girder. 

THE    EQUATION    OF    THE    ELASTIC    LINE. 

In  order  to  apply  equation  (l)  to  the 
case  of  continuous  girders,  we  have  to 
insert  for  M,  E  and  I  their  values  as  func 
tions  of  x  and  integrate  the  equation 
twice.  E,  however,  cannot  under  any 
ordinary  circumstances  be  a  function  of 
x,  it  being  dependent  upon  the  elasticity 
of  the  material  alone,  which  is  nearly  the 
same  in  one  and  the  same  beam,  and  we 
hence  regard  it  as  constant.  In  a  beam 
of  uniform  section  I  is  constant,  and 
although  it  varies  in  common  bridge 

o  o 

trusses,  we  shall  be  obliged  in  order  to 
bring  the  investigation  within  the  limits 


55 


of  this  paper  to  consider  it  always  as  con 
stant,  taking  care  to  point  out  after 
wards  the  slight  error  thus  introduced. 
Inserting  then  in  (1)  the  value  of  M 
from  (2),  we  have 


as  the  differential  equation,  applicable 
to  girders  of  constant  elasticity  and 
uniform  cross  section.  Integrating  this, 
the  constant  is  tr,  the  tangent  of  the  in 
clination  of  the  elastic  line  at  the  sup 
port  r  and 

dy  2Mrx-Sra 

*+  -- 


we  have  thus  far  taken  no  account  of 
the  relative  heights  of  the  supports. 
For  the  reasons  mentioned  in  the  last 
chapter,  we  shall  consider  them  as  all 
upon  the  same  level.  The  constant  for 
the  second  integration  is  then  0,  the 
origin  being  at  r,  and  we  have 

3  M   x2-S   x*  +  Prx 


56 

as  the  equation  of  the  elastic  curve  be 
tween  the  load  Pr  and  r-flth  support 
(Fig  5).  If  there  be  several  loads  we 
have  only  to  affix  the  sign  of  summation 
2  to  the  term  involving  Pr>  and  if  that 
term  be  omitted  we  shall  have  the  equa 
tion  between  the  load  and  the  rih  sup 
port,  since  for  any  section  between  those 
points  M=Mr—  Sr#. 

If  in  (5)  we  make  #—  /r,  y  becomes  0, 
and  inserting  for  Sr  its  value  from  (3), 
we  find 

(6)  6EH=-2Mrt 


Thus  the  equation  of  the  curve  is 
completely  determined,  when  we  know 
Mr  and  Mr  +  1  the  moments  at  the  sup 
ports  r  and  r-f-1.  These  may  be  found 
by  the  remarkable  theorem  of  three  mo 
ments. 

THE  THEOREM  OF  THREE  MOMENTS. 

In  Fig.  5  is  represented  a  portion  of  a 
continuous  truss.  Beginning  at  the  left 
hand  end,  the  lengths  of  the  spans  are 


57 

denoted  by  l^  /2,  .....  lr,  etc  ,  and  the  sup 
ports  are  designated  as  1,  2,  .....  r,  etc. 
Upon  the  spans  lr  —  \  and  /r  are  loads 
Pr  —  i  and  Pr>  whose  distances  from  the 
nearest  left  hand  supports  are  /£/r  —  i  and 
Jc  l^  k  being  any  fraction  less  than  unity, 
and  not  necessarily  the  same  in  the  two 
cases.  The  equation  of  the  elastic  line 
between  Pr  and  the  support  r  +  1  is 
given  by  (5),  and  the  tangent  of  the 
angle  which  the  curve  at  the  section  x 
makes  with  the  axis  of  abscissae  is  iven 

(4). 

Fio.  5. 

•^  ----  if-i  ~"x  -----  it-'' 


If  in  (4)  we  substitute  for  Sr  its  value 
from  (3),  and  for  tr  its  value  from    (6), 

and  make  cc=/r,  -~     becomes    t^\\    the 
ax 

tangent   of  the  inclination  of  the  curve 
at  r-f-1,  and  we  find 


58 

Now,  if  we  consider  the  origin  moved 
from  the  support  r  back  to  r—  1,  we  may 
derive  a  value  for  tr  by  simply  diminish 
ing  each  of  the  indices  in  the  above  ex 
pression  by  unity,  hence 

(7)  6  EH=Mr_i?r_i  +  2'Mr  /r_i 

-Pr_!Z2r  _!(&-&') 

Equating  the  values  of  6  E  I  tr  given   by 
(6)  and  (7),  we  have 

(8)  Mr_1/r_ 


which  is  the  most  general  form  of  the 
theorem  of  Three  Moments  for  girders 
of  constant  cross-section.  By  prefixing 
the  sign  2  to  the  terms  in  the  second 
member,  it  becomes  applicable  to  any 
number  of  single  loads.  For  uniform 

c_> 

loads  ^r  —  i  and  wr  per  linear  unit,  we 
have  only  to  place 

Pr  _  i  =  wr  —  id(klr  —  i)  =  wr  —  ilr  —  \dk 
Pr  =  wr  d  (k  lr)=ior  lr  d  k 

And  to  replace  the  sign  of  summation 
2  by  that  of  integration  /.  If  these 


59 


loads  extend  over  the  entire  spans  lr  and 
/r  —  1>  we  take  the  integrals  between  the 
limits  k=Q  and  &=1  and  have 


r  (/r    +    ?r  -  l)  +  Mr  +  l  1T 
W  _     13   _  WT    Is  T 


which  is  the  theorem  as  first  announced 
by  Clapeyron. 

For  each  support  of  a  continuous 
girder  an  equation  may  be  therefore 
written  involving  the  moment  at  that 
support,  and  those  at  the  preceding  and 
following  support.  In  a  girder  of  s 
spans  there  are  5+1  supports,  and  since 
the  moments  at  the  first  and  last  sup 
ports  are  zero  we  have  s  —  1  moments, 
whose  values  may  be  found  by  the  so 
lution  of  the  5—1  equations.  The  mo 
ments  give  the  shear  at  any  support,  and 
by  (2)  the  internal  forces  or  strains  may 
be  determined  for  every  section  of  the 
girder. 

REMARKS  ON  THE  PRECEDING  THEORY. 

The  laws  of  the  theory  of  continuity 
above  deduced  can  be  regarded  as  onlyap- 


60 

proximate.  Of  the  three  hypotheses  upon 
which  the  differential  equation  of  the 
elastic  line  is  deduced,  the  first,  although 
a  most  reasonable  assumption,  has  not 
been  definitely  verified  by  experiment, 
and  the  second  is  rendered  somewhat 
doubtful  by  the  extreme  difficulty  in 
delicate  experiments  of  assigning  the 
elastic  limits.  Nevertheless  they  are 
universally  regarded  by  all  writers  as 
sufficiently  accurate  to  form  the  basis  of 
a  working  theory,  and  must  continue  to 
be  thus  used  until  we  attain  to  a  more 
thorough  knowledge  of  matter  and  force. 
The  third  hypothesis,  however,  is  a  limi 
tation  of  the  data,  which  we  are  at  per 
fect  liberty  to  make,  since  we  know  that 
the  increase  in  length  of  the  girder  by 
deflection  is  too  small  to  be  practically 
measured.  We  may  conclude  then  that 
the  equation  (1)  is  an  extremely  close 
approximation  to  the  actual  law  govern 
ing  straight  elastic  beams.  From  the 
time  of  Navier  to  the  present  it  has 
been  so  accepted  and  used. 

The   next  hypothesis  or  limitation  of 


61 

data,  which  we  make,  is  that  E,  the  co 
efficient  of  elasticity,  is  constant  through 
out  the  girder.  In  a  solid  beam  of 
ordinary  homogeneous  material,  there 
can  be  no  reason  for  supposing  it  other 
wise. 

In  the  Journal  of  the  American  So 
ciety  of  Civil  Engineers  for  May,  1876, 
appeared  an  article  by  Charles  Bender, 
C.  E.,  in  which  the  use  of  continuous 
bridges  is  strongly  opposed.  One  of  his 
main  objections  is — that  the  theory  upon 
which  such  bridges  are  computed  is  un 
reliable  in  consequence  of  the  assump 
tion  of  a  constant  coefficient  of  elasticity, 
and  he  quotes  the  records  of  experi 
menters  to  show  that  values  for  the 
coefficient  of  iron  and  steel  have  been 
observed  ranging  from  17,000,000  to 
40,000,000  pounds  per  square  inch. 
These  limiting  values  are,  however, 
decidedly  exceptional,  but  even  grant 
ing  that  such  variations  may  exist 
in  materials  and  forms  like  soft  iron 
wire,  steel  rails  and  wrought  iron  eye- 
bars,  it  cannot  be  supposed  that  they 


62 

will  occur  in  one  and  the  same  structure, 
where  the  material  is  of  one  kind,  of 
similar  cross  sections  and  which  has 
been  subjected  in  the  same  mill  to  the 
same  process  of  manufacture.  The  mere 
statement  that  Morin  has  observed  val 
ues  of  the  coefficient  of  elasticity  as  low 
as  17,000,000  has  very  little  weight  when 
unaccompanied  by  any  reference  to  the 
kind  of  iron  experimented  upon.  Let 
us  see  what  Morin  himself  actually  says 
in  recapitulating  the  results  of  experi 
ments  upon  wrought  iron.* 

"  Iron  of  superior  quality,  which  comes 
from  standard  ores,  and  which  has  been 
manufactured  exclusively  with  charcoal, 
or  iron  from  sheet  metal,  many  times 
refined,  may  give  for  the  coefficient  of 
elasticity  values  as  high  as  E=28,- 
400,000,  or  even  E= 31, 200,000  Ibs.  per 
square  inch,  equal  to  those  furnished  by 
ordinary  steel.  Iron  of  ordinary  manu 
facture  reduced  with  common  coal,  and 
drawn  into  forms  like  rails,  T  irons, 

*Morin  ;    Resistance  des  Materiaux,  Paris,  1862.   Vol. 
I,  p.  443. 


63 

flanges,  etc.,  give  such  values  as 
E  =  24,100,000  and  E  =  '25,600,000. 
Finally  the  most  soft  and  ductile  iron 
furnishes  values  as  low  as  E=21, 300,000, 
or  E=  19,800,000,  or  even  E=:  17,000,000. 
It  is  well,  then,  in  calculations  upon  the 
strength  of  iron,  to  ascertain  the  quality 
of  the  material  and  the  process  of  manu 
facture." 

Mr.  Bender  likewise  alludes  to  ex 
periments  upon  wrought  iron  bars  in 
which  the  coefficient  of  elasticity  was 
found  to  be  40,000,000,  and  it  seems  to 
be  implied  by  his  language  that  such 
values  are  of  common  occurrence.  The 
fact,  however,  that  a  standard  writer  on 
the  strength  of  materials,  like  Morin, 
regards  34,000,000  as  an  exceptionally 
high  figure  for  iron,  may  justify  us  in 
demanding  that  when  a  value  like 
40,000,000  is  quoted,  we  should  be  fur 
nished  with  some  details  concerning  the 
quality  of  such  iron,  the  process  of 
manufacture,  as  well  as  a  description  of 
the  testing  machine  and  the  manner  of 
measuring  the  small  extensions  or  com- 


V  64 

pressions,  from  which  the  coefficient  is. 
calculated,  or  at  least  that  we  should  be 
referred  to  the  book  or  journal  where 
such  experiments  are  described.  And 
further  as  it  is  well  known  that  by  strain 
ing  a  bar  beyond  the  elastic  limits,  very 
low  values  of  E  can  be  deduced,  are  we 
not  justified  in  asking  similar  informa 
tion  concerning  experiments  which  fur 
nish  such  values  ? 

Undoubtedly  there  is  some  variation 
in  the  elasticity  of  different  pieces  of 
iron,  even  when  great  care  has  been 
taken  to  ensure  uniformity  of  material 
and  manufacture,  and  it  is  greatly  to  be 
desired  that  experiments  should  be  made 
to  determine  how  it  varies  with  the 
cross  section  and  length  of  the  piece. 
As  soon  as  such  a  law  of  variation  is 
discovered  (if  any  exist),  we  shall  be 
obliged  to  consider  E  as  variable  in  in 
vestigating  a  continuous  truss.  But  if 
no  law  exists  and  we  know  only  the  fact 
that  there  are  slight  variations  in  the 
elasticity  of  different  pieces  in  the  same 
truss,  what  is  to  be  done  ?  Nothing  but 


65 


to  regard  E  as  constant, 
assured  that  the  distribution  of  the 
variable  pieces  throughout  the  structure 
will  be  governed  by  the  law  of  proba 
bility,  and  that  hence  the  girder  as  a 
whole  will  conform  closely  to  the 
theoretic  form  of  the  elastic  line. 

The  next  argument  which  it  is  our 
duty  to  criticise  in  that  article  is,  that 
the  theory  of  continuous  girders  is  un 
reliable,  because  the  calculated  deflection 
does  not  generally  agree  with  the 
actually  measured  deflection.  The  ac 
curacy  of  the  computed  strains  must 
depend  upon  the  accuracy  of  the 
theorem  of  three  moments,  and  this  it  is 
asserted  depends  upon  the  calculated 
deflection.  And  because  the  deflection 
as  actually  measured  is  sometimes  no 
more  than  one-half  of  the  calculated 
one,  hence,  it  is  said,  the  same  differences 
may  occur  in  the  strains,  and  the  whole 
theory  is  unworthy  of  consideration. 

This  we  can  only  regard  as  a  striking 
instance  of  the  jncompetency  of  prac 
tical  men  to  draw  conclusions  from  even 


66 


simple  experiments.  The  reader  who 
has  followed  onr  presentation  of  the 
theory  of  the  elastic  line,  will  see  at 
once  that  the  value  of  the  deflection 
given  by  (5)  only  enters  the  discussion 
as  an  auxiliary  for  finding  (6),  the 
tangent  of  the  inclination  angle  at  the 
support.  The  process  supposes,  indeed, 
that  E  is  constant,  but  it  supposes  noth 
ing  whatever  concerning  the  value  of 
the  deflection  at  any  point.  When  we 
pass  to  the  next  span  and  find  again  in 
(7)  a  second  value  of  the  tangent,  the 
actual  value  of  the  deflection  there  is 
likewise  not  considered.  And  when  by 
the  combination  of  (6)  and  (7),  we  de 
duce  (8)  in  which  E  does  not  appear,  its 
very  absence  is  a  proof  that  the  moments 
and  hence  the  strains  are  entirely  in 
dependent  of  its  value  or  of  the  actual 
deflection.  If  two  trusses  of  the  same 
spans,  height  and  form  are  continuous 
over  several  supports,  one  of  steel  hav 
ing  a  coefficient  of  elasticity  of  31,000,000 
Ibs.  per  square  inch,  and  the  other  of 
wood  having  a  coefficient  of  only  one- 


67 

twentieth  as  much,  the  reactions,  shears, 
moments  and  strains  would  be  the  same 
in  each.  The  measurement  of  the  actual 
deflections  of  these  bridges  under  given 
loads  is  only  useful  for  determining  and 
comparing  their  stiffness  or  elasticity,  or 
in  connection  with  theory  for  finding  the 
values  of  E  and  I.  The  theory  of  con 
tinuity  rests  not  upon  absolute  deflec 
tions,  but  on  relative  ones— on  the  form 
of  the  elastic  curve,  and  this  again  upon 
the  three  universally  accepted  hypotheses 
included  in  our  equation  (1),  with  the 
additional  assumption  that  the  coefficient 
of  elasticity  is  practically  constant. 

One  more  remark  and  we  close  for  to 
day  a  discussion  which  shall  be  resumed 
in  our  next  chapter.  Mr.  Bender  advises 
us  to  abandon  the  theory  of  flexure,  to 
make  no  further  advance  in  bridge 
building,  to  remain  content  with  the 
simple  lever,  or  at  the  utmost  with  the 
continuous  (sic)  patent  hinged  truss. 
But  until  such  advice  is  enforced  by 
more  logical  arguments  than  we  have 
yet  seen,  we  must  continue  our  work 


68 


in  support  of  a  theory  and  system 
which  is  universally  accepted  as  only 
slightly  deviating  from  the  exact  ex 
isting  conditions,  which  is  applied  in 
the  erection  of  continuous  bridges  by 
every  nation  except  our  own,  and  which 
perhaps  if  carried  out  by  us  might  lead 
to  more  perfect  structures  than  the 
world  has  yet  seen.  The  great  majority 
of  coefficients  of  elasticity  quoted  in  his 
paper,  made  by  such  men  as  Staudinger, 
Baker,  Morin  and  Wohler,  were  in  fact 
found  by  measuring  the  deflections  of 
beams,  and  then  from  the  theory  of 
flexure  computing  the  value  of  E.  He 
accepts  those  values,  and  on  their  evi 
dence  condemns  the  theory  by  which 
they  were  deduced  !  Is  not  Morin's 
conclusion,  which  we  have  quoted  above, 
by  far  the  most  reasonable? 

MOMENTS    AT    THE    SUPPORTS. 

The  theorem  of  three  moments  given 
by  (8)  furnishes  the  means  of  finding  the 
moments  at  the  supports  due  to  any  as 
signed  system  of  loading.  The  actual 


69 

solution  of  those  equations  is,  however, 
quite  tedious  when  the  number  of  spans 
is  large,  and  we  proceed  therefore  to  de 
velop  a  general  solution,  by  which  the 
values  of  the  moments  may  be  formulated 
and  placed  in  a  convenient  form  for  nu 
merical  computation. 

In  the  designing  of  continuous  bridges 
it  is  only  necessary  to  consider  single 
loads  concentrated  at  the  panel  apices  or 
uniform  loads  extending  over  an  entire 
span.  Let  us,  then,  consider  a  continu 
ous  girder  of  constant  cross-section  and 
homogeneous  material  whose  supports 
are  on  the  same  level.  Let,  as  in  Fig. 
6,  the  supports  beginning  at  the  left 
hand  end  be  designated  by  the  indices  1, 
2,  3, ....  r,  etc,  and  the  lengths  of  the 
spans  Zt,  l#  l» ln  etc.  Call  the  num 
ber  of  spans  s;  then  the  last  span  will 
be  /„  and  the  last  support  s+l.  The 
ends  of  the  girders  rest  upon  abutments 
in  the  usual  manner,  the  lengths  of  the 
spans  may  be  all  different  and  their 
number  may  range  from  one  to  infinity. 
In  the  span  1T  let  a  single  load  P  be 


70 


placed  at  a  distance  k  1T  from  the  sup 
port  r,  (k  being  any  fraction  less  than 

FIG.  6. 


unity),  or  let  this  span  be  loaded  uni 
formly  from  r  to  r+  1  with  a  weight  W, 
(W  being  equal  to  w  /r,  if  w  is  the  load 
per  linear  unit),  all  the  other  spans  be 
ing  unloaded.  By  reference  to  (8)  we 
notice  that  there  are  two  functions  of 
such  loads  which  enter  the  equations  of 
moments.  If  the  single  load  P  is  alone 

V--- 

considered  these  functions  are 


the  first  entering  into  the  equation  for 
the  preceding  support  r,  and  the  second 
into  the  one  for  the  following  support 
r  4-  1.  If  the  uniform  load  over  the  whole 
span  is  alone  considered,  these  become 
each  equal  to  J  W  C,  as  we  have  shown 
above  in  discussing  the  theorem  of  three 


71 

moments.      In  the  following  investiga 
tion,  we  place  therefore  for  abbreviation 


A=P  5  (2*—  3  **  +  *')  )     for  a  single 
B=P  II  (*—  i*)  j  load  in  span  I,. 

(I) 

A—  \  -WZJ   )  for  a  uniform  load  over 
B=J  WZ*    f  whole  span  Zr. 

By  introducing  the  letters  A  and  B  to 
represent  these  functions,  our  discussion 
will  apply  equally  well  to  a  single  load 
P,  or  to  a  weight  W  uniformly  distribut 
ed  over  the  whole  of  a  single  span. 

Since  the  girder  is  not  fastened  at  the 
abutments  1  and  s-fl,  the  moments  at 
those  points  will  be  zero.  The  moments 
at  2,  3,  ....  r,  etc.,  we  designate  by  M2, 
M3,  ....  Mr,  etc.,  and  from  (8)  we  may 
write  an  equation  for  each  of  those  sup 
ports.  As  there  is  no  load  considered 
except  on  the  span  Zr,  the  right  hand 
member  of  the  equation  for  the  support 
r  will  be  A,  of  that  for  r  +  1  will  be  B 
and  of  all  the  others  will  be  zero.  Thus 
we  have  the  following  equations  : 


>7f 


2 


t 
The  number  of  these  equations  is  s—  1, 

the  same  as  the  number  of  unknown  mo 
ments.  Their  solution  is  best  effected 
by  the  method  of  indeterminate  multi 
pliers.  Let  then  the  first  equation  be 
multiplied  by  a  number  c2,  the  second 
by  ca,  etc.,  the  index  of  the  as  yet  inde 
terminate  numbers  corresponding  with 
that  of  the  M  in  the  middle  term.  Then 
let  all  the  equations,  thus  multiplied,  be 
added,  and  the  resulting  equation  be  ar 
ranged  according  to  the  coefficients  of 
the  unknown  moments  M2,  M3,  etc.  Now, 
if  we  require  that  such  relations  exist 
between  the  multipliers,  that  all  the 
terms  in  the  first  member  shall  reduce  to 


zero,  except  the  last  containing  M8>  the 
value  of  M8  is 


M= 


^I/S^JL+SA  (/8-i  K) 

And  the  values  of  the  multipliers  will 
be  given  by  the  equations 


etc.,         etc 

After  deducing  the  values  of  c  from 
these  equations,  the  value  M8  is  at  once 
known. 

Again,  if  we  multiply  the  equations  of 
moments,  beginning  with  the  last,  by  the 
indeterminate  numbers  d»  d3,  etc.,  all  the 
moments  except  M2  may  be  eliminated, 
and  we  have 


and  the  multipliers  will  bo  given  by  the 
equations 


etc.,  etc.,  etc. 


74 

The  values  of  the  numbers  in  the  series 
c  and  d  need  only  satisfy  the  equations 
as  given  above.  Assuming  then  c,=l, 
and  d,—  1,  we  get  the  following  : 

c,  =  0 
<V=l 

2A+A 

C3_  ^ 


(II)  Ci=-2c3 

3 

c.=  -2c,-(2c4  +  c.)  J 

c6=+2c6-(2cB  +  c4)-^ 

6 

etc.,  etc. 


etc.,  etc. 


75 

which  reduce  to  numerical  form  as  soon 
as  the  lengths  of  the  spans  for  any  par 
ticular  case  are  substituted. 

Since  the  5—1  equations  of  moments 
are  of  the  same  form  as  the  equations  of 
the  multipliers  c  and  d,  we  must  have 


or,  if  n  indicate  the  index  of  any  sup 
port, 

Mn  =  cn  JVI2       when  n <r  -f 1 
Mn  =  ^s_n-f2  MS,  when  n>r 

Inserting  in  these  the  values  of  M2  and 
M8  as  found  above,  we  have 

Mn  = 
(III) 

Mn  = 


when 

which  give  the  values  of  the  moments  at 
all  supports  in  terms  of  the  quantities  A 
and  B,  depending  only  upon  the  charac 
ter  of  the  load  and  its  position  in  the 


76 

span  /r,  and  the  numbers  c  and  d  depend 
ing  only  upon  the  lengths  and  number 
of  the  spans  of  the  girder.*  To  find 
their  numerical  value  for  any  given  case 
is  hence  a  simple  arithmetical  exercise. 

Example  1.  —  A  continuous  girder  has 
four  unequal  spans,  ^  =  80  ft.,/2=100ft., 
13=  50  ft.,  and  /4=40  ft.  (Let  the  reader 
draw  the  figure).  On  the  span  Z2  is  a 
single  load  P=10  tons,  whose  distance 
from  the  support  2  is  &Z2=40ft.  To 
find  the  moments  at  the  supports. 

Since  /£/2  =  40,  and  £2=100,  we  have 
&=0.4.  Inserting  then  in  (I),  the  values 
of  &,  lz  and  P,  we  find 

A=38400  tons  ft.     B-  -33600  tons  ft. 

Inserting  next  the  lengths  of  the  spans 
in  (II),  we  have 

^=0  ^=0 

c,=  l  4,  =  ! 

c—  —  3.6  c?=:—  3.6 


*  In  the  London  Philosophical  Magazine  for  September, 
1875,  where  the  above  demonstration  was  first  give'",  the 
author  has-  extended  the  method  to  girders  witn  fastened 
or  walled-in  ends. 


77 

Since  the  load  is  in  the  second  span  r— 2, 
also  5=4;  hence  r/s_r_f_2=f/4=10.3,  cr  = 
£2=1,  etc.,  and  /8  _  ^^^SO,  etc.  By- 
inserting  these  values  of  c,  d,  I,  A  and  B 
in  (III),  we  obtain 

Mn  =      82.01  cn,         when  n<3 
Mn  =  --24.05  £/6-n,  when  n>2 

For  the  abutment  or  left  hand  support, 
we  have  n=lr  c,=0,  and  hence  IV^— 0, 
for  the  second  support,  n  =  2,  ^2=1,  and 
M2=S2.01  tons  ft.  For  the  third  sup 
port,  w=3,  f/6_nr=  —  3.6,  and  M3  =  88.56 
tons  ft.  For  the  next,  ?i=4,  r/e— n:=l> 
and  M3=  — 24.65  tons  ft.  Lastly,  for  the 
right  hand  abutment,  w  =  5,'and  M6— 0. 
A  positive  moment,  it  will  be  remember 
ed,  causes  tension  in  the  upper  chord  of 
a  truss,  while  a  negative  moment  causes 
the  reverse. 

2.  A  girder  of  four  equal  spans  has  a 
load  P  at  any  point  on  the  tirst  span. 
Find  the  moment  at  each  pier  due  to  P. 

Ans.  M4=if  PJ  (k— &3),  etc. 


78 

SHEARS  AND  REACTIONS  AT  THE  SUPPORTS. 

It  is  thus  easy  from  (I),  (II)  and  (III) 
to  find  the  moments  at  all  supports  due 
either  to  single  concentrated  loads  or  to 
a  uniform  load  over  an  entire  span,  and 
these  are  the  only  two  kinds  of  loading 
which  we  need  to  consider  in  designing 

O  O 

a  continuous  bridge.  We  next  need  the 
shear  Su  at  the  right  of  any  support  due 
to  these  same  loads. 

In  computing  strains  in  a  continuous 
truss  we  take  up  each  span  separately. 
The  index  n  refers  always  to  the  particu 
lar  span  under  consideration,  while  the 
index  r  referring  to  the  span  in  which 
the  load  is  for  the  moment  considered, 
may  be  less  than,  equal  to  or  greater 
than  n.  For  single  loads  the  shear  Sr  is 
given  directly  by  (3),  fora  uniform  load 
P  (1  —  k)  in  that  expression  becomes  ^  wl^ 
while  for  an  unloaded  span  those  terms 
disappear.  Thus  we  have 

-ft),  for  Fig.  6. 


(IV)  Sr  =*  for  Fig.  3. 

' 


79 
Mn-Mn 


,    whenn>r,  orn<r, 

&n 

for  the   shear  in  the   span   /n  infinitely 
near  to  the  support  n. 

The  shear  in  the  span  /n  at  a  point  in 
finitely  near  to  the  n+lih  support  is 
called  S'n  (see  Fig.  3).  For  its  values 
we  deduce 


- 

S'f  =     I-         +  P  k,  for  Fig.  6. 

' 


S'r  =     ll  +$  wfti  for  Fig.  3. 


r 


Q/  Mn  +  l—  Mn 

Sn  =  -  -,for  n>r,  or  n<r. 

'n 

The  reaction  I?n  at  any  support  n  is 
evidently  the  sum  of  the  shear  Sn  in  the 
span  ln  ;  and  of  the  shear  S'n_i  in  the 
span  /n—  i  O1"  f°r  all  cases 


Rn  =     n  +       n  —  !• 

Example.  —  A  girder  of  four  equal 
spans  has  a  single  load  P  at  the  center  of 
the  third  span  from  the  left  end.  Find 
the  reaction  at  the  third  support. 

Ans.  E  =  4JP. 


80 


SHEAR     AND     MOMENT     AT     ANY     SECTION. 

These  are  given  directly  by  the  sim 
ple  conditions  of  static  equilibrium. 
For  a  single  load  we  have  from  (2),  (see 
Figs.  5  and  6), 

S  =  Sr  —  P,  for  a  section  between 
(Y)  Pandr-fl 

S=Sn>  for  any  other  section. 

as  expressing  the  internal  shear  for  any 
section  x  (see  Figs.  5  and  6). 
Also,  we  have 

M=Mr  -  -  Sr  x  +  P  (x—klr\  between 

(VI)  Pandr-fl 

M=Mn  --  Snce,  for  any  other  section, 

for  the  internal  moment  for  any  section 
x  in  a  span  either  unloaded  or  containing 
the  weight  P.  Similar  expressions  may 
be  also  written  as  in  Chapter  I,  if  the 
load  be  taken  as  uniformly  distributed. 
JZxample. — A  girder  of  three  equal 
spans  has  a  load  P  at  the  center  of  the 
first  spans.  What  is  the  shear  and  mo 
ment  at  the  center  of  the  middle  span  ? 

Ans.  S=*P,  M=ftP/. 


81 

MAXIMUM  SHEARS  AND  MOMENTS. 

The  formulae  (I)  to  (VI)  inclusive  are 
sufficient  in  connection  with  an  arithmeti 
cal  process  of  tabulation  to  determine 
the  maximum  strains  in  all  the  pieces  of 
a  properly  designed  continuous  truss. 
Having  for  instance  to  calculate  tbe  span 
/n,  we  may  take  at  every  panel  apex 
throughout  the  bridge  a  single  load  P, 
and  compute  the  shear  and  moment  at 
any  section  due  to  every  possible  posi 
tion  of  P.  These  arranged  in  a  table, 
afford  a  clear  view  of  the  distribution  of 
loading  giving  the  maxima  ;  the  great 
est  positive  shear,  for  example,  occurring 
when  the  live  load  covers  those  portions 
of  the  bridge  which  furnish  plus  values 
of  S,  while  at  the  same  tim.3,  it  is  absent 
from  those  portions  giving  minus  values 
of  S.  Adding  then  all  the  plus  values 
thus  found,  the  maximum  is  determined 
by  combination  with  that  due  to  the  al 
ways  existing  dead  load. 

It  is  therefore  not  absolutely  neces 
sary  that  the  engineer  should  be  ac- 


82 

quainted  with  the  theory  of  the  distri 
bution  of  loading  giving  rise  to  the  maxi- 
murn  strains  in  the  various  pieces  of  the 
truss.  As  such  a  knowledge,  however, 
is  of  great  assistance  in  checking  the  ac 
curacy  of  the  calculation,  we  shall  here 
state  without  demonstration  the  cases 

under  which  such  maxima  and  minima 
i 

arise. 

First  the  shear  ;  from  this  we  obtain 
the  strains  in  the  webbing  by  the  simple 
multiplication  by  a  constant,  a  positive 
shear  producing  tension  in  a  diagonal 
which  slopes  upward  toward  the  left 
hand  support.  The  maximum  positive 
shear  in  the  span  lu  at  the  section  whose 
distance  from  the  support  n  is  cc,  occurs 
under  a  distribution  of  loading  such  as 
is  represented  in  Fig.  7,  in  which  the 
shaded  portions  denote  the  live  or  rolling 
load.  From  this  we  see  that  the  nearest 
span  on  the  left  and  each  alternate  one 

FJG.  7. 


u-/         n 


83 

are  covered  with  the  live  load  ;  that 
from  the  section  x  to  the  support  n-f-1 
the  live  load  extends  ;  and  that  the  sec 
ond  span  on  the  right  and  each  alternate 
one  are  also  covered  with  the  live  load  ; 
all  other  portions  being  subjected  only 
to  the  dead  or  actual  load  of  the  truss. 
The  minimum  positive,  or,  what  is  the 
same  thing,  the  maximum  negative  shear 
obtains  under  exactly  reverse  conditions, 
the  loaded  portions  in  Fig.  7  being  un 
loaded,  while  the  empty  ones  receive  the 
live  load.  Let  the  reader  draw  a  figure 
for  this  case,  and  imagine  the  section  x 
to  move  from  n  to  n-h  1. 

Next  the  moment  ;  from  this  we  ob 
tain  the  chord  strains  by  dividing  by  the 
constant  depth  of  the  truss,  a  positive 
moment  producing  tension  in  the  upper 
chord.  Here  the  maximum  positive  mo 
ment  in  the  span  /n>  occurs  near  the  sup 
port  n  under  a  distribution  of  loading 
like  that  represented  in  the  first  illustra 
tion  of  Fig.  8,  near  the  middle  of  the 
span  as  in  the  second  and  near  the  sup 
port  tt-fl  as  in  the  last.  Fig.  8  repre- 


84 


sents  one  and  the  same  beam  with  the 
cases  of  loading  causing  maximum  posi 
tive  moments  at  three  different  sections 
in  the  span  lUf  the  first  a  section  between 
n  and  a  point  i,  the  second  between  i 
and  i'  and  the  third  between  i'  and  w-f  1. 
These  points  i  and  i'  are  called  fixed  in 
flection  points,  and  they  enjoy  the  pro 
perty  that  all  loads  on  the  spans  to  the 

FIG.  8. 


—-Ln • 


71H        71' 


71*,' 


atb. 


Mb. 


right  of  lUt  produce  no  moment  at  i, 
while  all  loads  on  the  spans  to  the  left 
of  ln  produce  no  moment  at  i'.  The 
position  of  these  points  depend  only  up 
on  the  lengths  and  number  of  the  spans 


85 

of  the  girder,  or  upon  the  numbers  c  and 
d  given  by  (II).  If  the  distance  from  n 
to  i  be  denoted  by  i,  and  that  from  n  to 
ir  by  i'9  the  following  simple  formulae 

<?n  ni 


.,__          $s  — n-f-2  ^n 


— n-j-2  —  <    —  n-f  1 

will  give  the  position  of  the  fixed^inflec- 
tion  points  in  any  span  £n. 

In  order  to  render  these  distributions 
of  load  clear,  let  us  imagine  the  sec 
tion  x  to  move  from  the  support  n  to 
Ti+1.  When  the  section  is  at  n  the  live 
load  covers  the  whole  span  £n  to  render 
the  moment  at  a;  a  maximum,  as  x 
passes  toward  i  the  load  recedes  rapidly 
toward  n  +  1,  until  when  x  reaches  i  the 
span  ln  becomes  empty,  and  the  loads 
on  the  following  spans  shift  as  shown  in 
Fig.  8.  As  x  passes  from  i  to  ir  the 
span  lu  remains  empty  as  in  the  second 
sketch  and  when  it  reaches  i'  the  loads 
on  the  preceding  spans  shift.  As  soon 
as  x  passes  i'  the  load  begins  to  come  on 


86 

at  n,  which  rapidly  increases  as  x  moves, 
until  it  covers  the  whole  span  when  x 
coincides  with  n+l. 

The  arrangements  of  loading  for  caus 
ing  the  maximum  negative  moment  in 
any  section  depend  likewise  upon  the 
position  of  that  section  with  reference  to 
the  fixed  inflection  points,  and  are  in  all 
cases  exactly  the  reverse  of  those  for  the 
positive  moment. 

It  will  be  seen,  then,  that  the  maxi 
mum  moments  between  the  supports  and 
the  fixed  inflection  points  cannot  be  de 
termined  by  cases  of  loading,  for  such 
cases  are  different  for  every  section.  In 
a  giider  of  two  equal  spans  for  example, 
one  of  these  points  in  each  span  coin 
cides  with  the  abutments,  the  others  are 
at  one-fifth  the  length  of  the  span  from 
the  pier.  Between  those  points  the 
maximum  strains  are  not  to  be  found  by 
parabolic  curves  of  moments  drawn  from 
a  few  assumed  arrangements  of  loading. 
Here  have  some  late  writers  fallen  into 
grave  error. 

The  above  completes,  what  seems  to 


87 

us  a  simple  presentation  of  the  theory  of 
the  continuous  girder  of  constant  cross- 
section.  We  have  disconnected  it  en 
tirely  from  the  properties  of  the  simple 
girder,  have  avoided  the  use  of  artificial 
angles  and  couples,  parabolic  moment 
and  shear  curves,  static  and  elastic  reac 
tions  and  other  paraphernalia  which  are 
too  often  introduced  to  complicate  the 
subject.  The  formulae  (I)  to  (VI)  which 
may  be  written  on  a  page  of  the  engi 
neer's  note  book  include  indeed  the 
whole  theory,  and  are  sufficient  for  the 
determination  of  the  maximum  strains  in 
a  continuous  truss  of  any  number  or 
lengths  of  spans. 


89 


CHAPTER  III. 

We  will  now  apply  the  above  theory 
to  the  practical  calculation  of  the  strains 
in  a  continuous  trus«,  and  to  show  the 
perfect  generality  of  our  method  we  will 
take  one  oftftve  unequal  spans. 

Fig.  9  shows  the  relative  lengths  of 
the  several  spans,  each  support  and  span 
receiving  an  index  according  to  the  no 
tation  previously  adopted.  The  first 
span  Z,  has  a  length  of  70  feet,  the 

FIG.  9. 
It        72  £3  Iff-  15 

y  &       """nr"  ^       £& 

/       *  3         *  56 

second  l^  of  100  feet,  the  third  l%  of 
80  feet,  the  fourth  14  of  120  feet  and 
the  last  Jt  of  90  feet.  This  girder  is 


90 


to  be  subject  to  a  live  load  of  0.8  tons 
per  linear  foot  per  truss  ;  the  dead 
load  we  estimate  at  0.6  tons  per 
linear  foot  per  truss.  It  is  divided 
into  panels  of  ten  feet  each  and  its 
height  is  also  ten  feet,  the  webbing  being 
a  simple  series  of  isosceles  triangles  as 
shown  in  Fig.  10,  which  represents  the 

FIG.  10. 

<u I,  =  SO- , 

A       BC        D       E   3  r      is       H       tc      I 


span  13  enlarged.  The  live  load  is  ap 
plied  at  the  panel  points  on  the  lower 
chord.  It  is  required  to  calculate  the 
maximum  strains  in  all  the  pieces  of  the 
span  19  due  to  the  above  live  and  dead 
loads. 

We  take  up  first  the  live  load  of  0.8 
tons  per  linear  foot,  or  eight  tons  per 
panel.  Since  every  load  in  the  span  ^ 
affects  every  section  in  13  in  a  similar 
manner,  we  may,  instead  of  considering 


91 

the  separate  panel  loads  on  /„  take  them 
as  uniformly  distributed  in  the  prelimi 
nary  determination  of  the  shear  and 
moment  at  3.  (The  load  of  eight  tons  at 
the  points  1,  2,  3,  etc.,  give  reactions 
only  at  those  points,  and  cannot  affect 
the  span  Z3).  On  the  span  /,  there  are 
seven  panels  and  six  apices,  hence  the 
live  load  in  that  span  is  W,  =  6XS  =  48 
tons.  In  the  same  way  we  have  on  the 
spans  Ja,  /4  and  Z6,  to  consider  the  live 
loads  applied  at  the  panel  apices  as  uni 
formly  distributed  over  the  spans;  but  in 
the  span  13  we  must  consider  each  panel 
load  separately,  since  different  arrange 
ments  of  those  loads  give  maxima  for 
different  sections.  Thus  we  have 


On  119  the  load  W1  =  48  tons, 
On  ^,  the  load  Wa=72  tons, 
On  *3,  the  loads  P,,  Pu,  P3,  etc., 

(see  Fig.  10)  each  equal  to  8  tons, 

On  Z4,  the  load  W4=88  tons, 
On  Z6,  the  load  W5=64  tons. 

We  now  turn  to  formulae  (I)   of  the 


92 

preceding  chapter,  and  determine  the 
quantities  A  and  B  due  to  each  of  these 
loads.  For  that  on  /,  we  have,  for  ex 
ample,  Wt=48,  /\=:4900,  hence  A  =  B 
=58800.  For  Pt  we  have  P=8,  l\— 
6400,  k—%,  hence  A=  10500,  and  B= 
6300  ;  in  like  manner,  for  P2,  P3,  etc.,  we 
place  &=t,  ^^f,  etc.,  and  find  for  each 
a  value  of  A  and  B.  Thus  we  have  for 
the  several  loads  : 

For  Wt,  A=B=  58800 

For  W2,  A=B=  180000 

ForPj,  A=  10500     B=   6300 

ForP,,  A=16800     B=12000 

ForP3,  A=19500     B=16500 

ForP4,  A=19200     B  =  19200 

ForP6,  A— 16500     B  =  19500 

ForP6,  A-12000     B  =  16800 

ForP7,  A=   6300     B=  10500 

For  W4,  A=B  =  316800 

ForW6,  A=B=  129600 

We  next  turn  to  formulae  (II),  and  sub 
stitute  the  lengths  ^  =  70,  /,=  100,  etc., 
and  thus  obtain 


93 


c,=0  d, 


'*  —  "*  — 

»   — 04.  /7   —         Q   K 

3 ^'?  "3 ***9 


c=—  44.567         <—  —  54.8 

o  5 

as  the  values  of  the  multipliers  c  and  d 
for  the  case  under  consideration. 

We  are  now  able  to  find  from  (III) 
the  moments  Ms  and  M4  at  the  supports 
3  and  4  for  each  of  the  above  loads. 
Since  there  are  five  spans,  5—5,  dn= 
—54.8,  c78_i  —  16,  etc.  ;  substituting  these 
in  (III),  we  have 

r 
-,whenn<r-fl 


*r  cr-frr4_i 

Mn  =  -^T-n-  -,  when  n>r 

1  t  U  O  A 

n  being  any  index  (in  our  case  either  3 
or  4),  and  r  the  index  of  that  span, 
which,  for  the  moment,  we  regard  as 
loaded.  Taking  the  load  on  the  first 
span,  we  have  r=l,  and  since  n>r,  we 
use  only  the  second  of  the  above  formu 
lae,  which  becomes 


94 


A  c,  4-  B  <r 
Mn  -  -rfT_n_-i__  -  3.452  <7T_n 

Substituting  in  this  TI—  3  and  w—  4,  and 
we  have 

M3=  —  3.452  (16)  =  —  55.23  tons  ft. 
M4=  —  3.452  (—3.5)  =  12.08  tons  ft. 

Taking  the  load  in  the  second  span,  we 
have  r=2,  and 


-,.-  0          . 

Mn  =  -^7_n—  --3  - 


in  which,  by  making  n=3  and  n=4,  we 
get  the  values  of  M3  and  M4. 

For  the  single  loads  on  the  span  131  we 
must  use  the  first  of  formulae  (III)  to 
obtain  M3,  and  the  second  to  obtain  M4. 
Making  then  r—  3,  we  have, 

3.4 


3  17032  17032 


_  A  ^  +  B  C4         3.5 

M  ^2  —  //  — 

4 


(16A-3.5B) 


17032  17032 

(-3.4A  +  14.05  B) 

and  inserting  in  these  the  values  of  A 


and  B,  as  given  above,  we  find  the  mo 
ments  at  3  and  4  from  each  of  the  loads 
from  Px  to  P7.  For  the  load  on  the 
fourth  span,  we  make  r=4;  for  that  on 
the  fifth  span,  r=5;  and  the  first  of  for 
mulae  (III)  give  the  moments.  Thus,  by 
very  simple  arithmetical  work,  we  obtain 
the  moments  M3  and  M4  due  to  each  of 
the  single  loads  in  Z3,  and  each  of  the 
uniform  loads  in  the  exterior  spans,  and 
arrange  them  in  the  second  and  third 
columns  of  the  following  table  : 


Load 

M3 

M4 

83 

tons  ft. 

tons  ft. 

tons. 

w, 

-  55.23 

+  12  08 

—  0.841 

W2 

+40").  82 

-  t^.77 

-  6.182 

p, 

+  81.19 

h  10.83 

-  7.254 

P2 

4-  45.36 

-  22.85 

-  6.281 

-  50.85 

-  83.93 

-  6.211 

pi 

r  48.00 

-  41.92 

-  4  078 

p.. 

-3D.  15 

-  44.65 

-  2.9U 

PC 

~  2(5.64 

-  39.92 

+  1.834 

P7 

-  12.81 

-  25.85 

+  0.837 

W4 

-  158.10 

-653.34 

-10.143 

W8 

+  25.87 

-106.91 

+  1.660 

The   last  column  of  the  table,  which 
gives   the  shear  S3  in  the  span  13  at  a 


06 

point  infinitely  near  to  the  support  3,  is 
found  from  the  quantities  M3  and  M4  by 
means  of  formulae  (IV).  The  load  Ws, 
for  example,  gives  a  positive  moment  of 
405.82  tons  ft.  at  3,  and  a  negative  one 
of  88.77  tons  ft.  at  4.  From  the  last 
formula  of  (IV),  we  have  then 


80 


Also  for  the  load  P3  on  the  span  /3,  we 
have  P=8,  &=i,  and  from  the  first  for- 

/  o  / 

mulae  of  (IV) 

p  _  50.85-33.93         (1      ax      fl  91  ,  tnns, 
80 

and  in  the  same  way  the  other  shears  in 
the  last  column  are  computed.  All  of 
these  refer,  of  course,  only  to  the  live 
load  of  eight  tons  per  panel. 

We  are  now  ready  to  proceed  with  the 
computation  of  the  maximum  strains  in 
the  span  7S.  And  first  we  take  up  the 
webbing. 

The  maximum  strain  in  any  diagonal 
in  Fig.  10,  is  equal  to  the  maximum 


shear  for  that  section  multiplied  by  the 
secant  of  the  angle  between  the  diagonal 
and  a  vertical.  We  proceed  first  to  find 
the  maximum  shears. 

The  shear  at  any  section  due  to  the 
dead  load  is  constant,  increases  or  de 
creases  as  the  live  load  comes  upon  the 
bridge,  and  becomes  a  maximum  or 
minimum  under  certain  particular  distri 
butions  of  loading.  To  determine  these 

o 

it  is  only  necessary  to  tabulate  the  shear 
due  to  each  separate  load.  This  is  easily 
done  from  the  values  of  S3  and  the 
formulae  (V).  In  the  following  table  the 
vertical  column  headed  aB/>  includes  the 
shears  which  may  act  upon  the  diagonals 
a  B  and  B  b,  b  C  c  those  for  b  C  and  C  c, 
etc.  The  horizontal  column  numbered  1 
gives  then  the  shears  at  every  section 
due  to  the  live  loads  ;  the  load  W,  for 
example  producing  a  negative  shear  of 
0.84  tons  in  every  panel  or  S=S3,  the 
load  P3  giving  in  the  three  panels  on  its 
left  S  —  S.  =  +5.21  tons  and  in  the  five 

•5 

on  its  right  S  =  S3— P=  +  5.21  —  8  —  —  2.79 
tons.  A  mere  inspection  of  this  table 


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shows  the  distributiorr  of  live  load  caus 
ing  the  maximum  or  minimum  shear  in 

c_? 

any  section.  Thus  for  the  panel  c?Ee 
the  maximum  occurs  when  those  loads 
giving  positive  shears  are  present, 
namely,  W,,  P4,  P.,  P.,  P,.  and  W.  and 
when  all  the  others  are  absent,  and  the 
minimum  occurs  when  only  those  giving 
negative  shears  are  on  the  bridge.  If 
then  we  add  all  the  positive  quantities 
in  1  and  likewise  all  the  negative  ones 
and  place  the  results  in  the  horizontal 
column  2,  we  have  for  the  panel  dl&e, 
4-17.52  tons  and  -16.24  tons  as  the 
greatest  and  least  shears  which  can  occur 
in  that  panel  due  to  the  live  load,  and 
these  need  only  to  be  combined  with  the 
shear  due  to  the  dead  load  to  obtain  the 
absolute  maximum  and  minimum. 

If  the  dead  load  be  regarded  like  the 
live  load  as  concentrated  at  the  panel 
points  on  the  lower  chord,  its  effect  will 
be  a  fractional  part  of  that  of  the  live 
considered  as  uniformly  distributed. 
Adding  algebraically  the  quantities  in  2 
we  have  in  3  the  shears  produced  by  a 


100 

uniformly  distributed  live  load  of  eight 
tons  per  panel,  since  this  is  the  same  as 
taking  the  algebraic  sum  of  all  the  quan 
tities  in  1.  The  live  load  if  extending 
over  the  whole  bridge  will  then  produce 
in  dTSe  a  shear  of  4-1.28  tons,  and  since 
the  actual  dead  load  is  three-fourths  of 
the  live,  the  dead  load  must  produce  in 
ttuit  panel  a  shear  equal  to  fxl.28 
=  0.96  tons.  Taking  then  three-fourths 
of  the  quantities  in  the  horizontal  col 
umn  3  we  have  in  4  the  shears  due  to 
the  dead  load  of  six  tons  per  panel. 

The  shears  in  4  always  must  exist, 
while  those  in  2  may  exist  under  certain 
positions  of  the  live  load.  The  absolute 
maxima  arc  therefore  found  by  adding 
algebraically  the  quantities  in  those  two 
horizontal  rows.  Thus  for<#Ee,  -f-0.96 
always  obtains,  and  if  +17.52  also  oc 
curs  their  sum  4-18.48  is  the  positive 
maximum  shear  ;  and  if  -16.24  occurs, 
4-0.96  —16.24  =  —  15.28  is  the  minimum 
or  negative  maximum.  Placing  these 
results  in  column  5  we  have  the  required 
maximum  shears  in  every  section  of  the 


101 

span  under  consideration.  If  the  dead 
load  shear  in  4  is  greater  than  the  live 
load  shear  of  opposite  sign  in  2  only  one 
kind  of  shear  can  prevail  ;  thus  in  bCc 
the  greatest  possible  value  is  +12.96 
+  29.01  =+41.97  tons,  the  least  possible 
is  +12.96  --11.73  =  +  1.23  tons  and  the 
diagonals  b  C  and  Cc  will  be  subject  to 
only  one  kind  of  strain.  In  the  case  be 
fore  us  three  panels  cDr7,  c/Ee  and 
e~Ff  have  both  a  positive  and  negative 
maximum  and  the  diagonals  in  those 
panels  may  be  subject  to  either  tension 
or  compression. 

The  maximum  shears  in  5  multiplied 
by  sec.  9,  or  the  secant  of  the  inclination 
of  diagonal  to  vertical  give  the  maximum 
strains,  tension  of  the  diagonal  slopes  up 
ward  toward  the  left  hand  support,  com 
pression  if  it  slopes  downward.  The 
depth  of  the  truss  being  ten  feet  and  the 
half  panel  length  5  feet,  sec.  6  is  1.118. 
We  have  then  the  following  table  of 
maximum  strains  in  the 


102 


DIAGONALS.     (See  Fisr.  10.) 


Ba 
B6 
Od 
Cc 
DC 
T>d 
Ed 


HA 
KA 
JLk 


— 61.7  tons 

+61.7 

—46.9 

+46.9 

2  or  +7.2  tons 
2  or  —7.2 

—20.6  or  +17.1 

+20  6  or  —17  1 

+28.2  or  —9.4 

—28.2  or  +9.4 

+40.5 

—40.5 

+.",4.0 

—54.0 

+688 

—68.3 


in  which  +  denotes  tension  and  —  com 
pression. 

In  the  same  manner  we  may  tabulate 
the    moments   at   every    section   due  to 

P 

each  load  and  deduce  the  maximum 
chord  strains.  For  the  upper  chord 
panels  A  B,  B  C,  C  D,  etc.,  the  centers 
of  moments  are  at  the  opposite  vertices 
a,  b,  c,  etc.,  and  hence  in  formulae  (VI) 
we  must  take  0,  10,  20,  etc.,  as  the  suc 
cessive  values  of  x.  To  find  the  moment 
for  CD  due  to  the  load  W,  on  the 


103 

span  £,,  we  have  only  to  insert  #— 20, 
and  the  values  of  M3  and  S3  as  found 
above  for  that  load  (see  Fig.  10),  giving 

M  =  —55.23  -  •  (—0.841  X  20) 

=  --  38.41  tons  ft. 

the  negative  sign  denoting  that  W, 
causes  compression  in  the  upper  chord. 
Also  to  find  the  moment  in  the  same  panel 
due  to  the  load  P  ,  we  have  as  before 
z=:20  and 

M  =  31.19—7.254X20  +  8X10 

=  —  33.89  tons  ft. 

In  this  way  we  readily  compute  the  mo 
ments  for  every  panel  due  to  every  live 
load  and  arrange  them  as  in  the  follow 
ing  table  of 


MOMENTS  FOR  UPPER  CHORD.  (See  Fig.  10.) 

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+ 

CO 

CO  O7 

+ 

o* 
n 

y 

r 

}O^ 

•I  1C  CO  O7  GO  00  ^  t 

-  CO 

0  t> 

+ 

+ 

0 

r  -^  ^ 
CO 

H    t< 

-    -- 

'  I— 

•(  t>  C5  GO  **  C 

5  C5 

66  c: 

CC  r- 

: 

- 

PQ 

O7  GO  O7  ^  3" 

5  O  07  ^ 

5  Xr- 

•<  d 

CO  CO 

07 

i- 

CO 

CO 

T-t 

0 

o 

TH 

O  »O  rH  JO  O  GO  Si  O  O7  y 

»C  O  CO  ^n"  1C  ^r  CO  O7  '—  '  i' 

zjL-H-t-H-H-- 

J  l!t 

ft 

1C  C' 
•y^  — 

CO  O' 

• 
• 

> 

SV---v>V 

-~A— 

>  ^ 

' 

' 

l~> 

CC 

a 

rt 
O 

+  1 

T-I 

05 

CO 

!"* 

0 

105 

The  horizontal  column  1  of  this  table 
shows  at  a  glance  the  distribution  of  live 
load  giving  the  maximum  strain  in  any 
bay  :  in  the  bay  B  C  for  instance  the 
loads  W2  W5  and  P4  to  P7  inclusive 
produce  positive  moments,  and  hence 
the  greatest  tensile  strain  obtains  when 
those  loads  are  on  the  bridge  and  all 
others  are  absent,  while  the  least  tensile 
strain  in  B  C  occurs  when  W1?  W4,  P,,  P, 
and  P3  are  present  and  the  others  ab 
sent.  Adding  separately  therefore  these 
positive  and  negative  moments,  we  have 
in  2,  the  greatest  and  least  moments  for 
every  bay  due  to  the  live  load.  Adding 
those  algebraically  and  we  have  in  3  the 
moments  when  the  live  load  covers  the 
entire  bridge ;  taking  three-fourths  of 
the  quantities  in  3  we  have  in  4  the  mo 
ments  due  to  the  actual  dead  load. 
Lastly  combining  the  moments  in  4  which 
always  must  exist  with  those  in  2  which 
may  exist,  we  get  in  5  the  absolute 
maxima  and  minima.  For  example,  in 
CD  we  have  due  to  the  live  load  the 
moments  +327.0  and  -281.3,  the  sum 


106 

of  these  or  +45.7  is  the  moment  when 
the  live  load  extends  over  the  whole 
girder,  three-fourths  of  this  or  -f  34.2  is 
the  value  due  to  the  actual  dead  load, 
and  finally 

+  34.2 -f  327.0=  -f- 361.2  tons  ft. 
+  34.2  —  281.3:^—247.1  tons  ft. 

are  the  maximum  positive  and  negative 
moments,  C  D  may  then  be  subject  to 
two  kinds  of  strain. 

Dividing  these  results  by  the  depth  of 
the  truss  and  remembering  that  a  posi 
tive  moment  causes  tension  in  the  upper 
chord,  we  have  the  maximum  strains  for 
the 

UPPER  CHORD.     (See  Fig.  10.) 


AB 

-104.0  tons 

BG 

-  547 

CD 

-  36  1  or  — 

24.7  tons 

DE 

-  33.2  or  — 

44  9 

EP 

-  30.1  or  — 

41  I 

FG 

-  4~>.0  or  — 

43.3 

GH 

-  (>0.2  or  — 

21.4 

Hit 

-  860 

KL 

-140.3 

107 

From  this  we  see  that  the  whole  upper 
chord  may  under  certain  positions  of  the 
rolling  load  be  subject  to  tension.  This 
is  due  to  the  short  length  of  the  span 
compared  with  the  adjacent  ones. 

The  calculation  of  the  maximum 
strains  in  the  lower  chord  is  entirely 
similar,  the  centers  of  moments  being  at 
the  opposite  vertices  B,  C,  D,  etc.,  and 
the  corresponding  values  of  x  being  5, 
15,  25,  etc.  We  leave  therefore  as  an 
exercise  for  the  student  the  formation  of 
the  tabulation,  merely  giving  the  results 
to  which  it  will  lead,  viz. 

LOWER  CHORD.     (See  Fig.  10.) 


ab 

—  76.9  tons 

be 

—  41.4  gry'  "+"  ^"»l 

cd 

—  34.7  or  - 

-34.  6  tons 

de 

—  84.7  or  - 

-478 

ef 

—  40.7  or  - 

-47.1 

fg 

—  51.9  or  - 

-34.0 

gh 

—  70.7 

lik 

—127.9 

The  strain  sheet  for  the  span  lz  (Fig.9) 
is  now  finished,   and  in  a   similar  way 


each  of  the  other  spans  may  be  comput 
ed.  The  method  we  have  presented  is 
entirely  general  and  applicable  to  any 
number  of  continuous  spans,  whether 
equal  or  unequal.  In  each  case  we  take 
the  load  in  the  exterior  spans  as  uniform 
and  that  in  the  span  under  consideration 
as  applied  at  the  panel  apices,  and  find 
for  each  the  quantities  A  and  B  from  (I) 
and  from  the  lengths  of  the  spans  we 
find  by  (II)  the  multipliers  c  and  d. 
These  enable  us  to  deduce  from  (III) 
the  moments  at  the  supports  due  to  each 
load,  from  which  (IV)  give  us  the  shear. 
It  is  only  in  this  preliminary  computation 
of  moments  and  shears  that  the  calcula 
tion  of  continuous  girders  differs  from 
that  of  ordinary  simple  trusses.  In  the 
latter  the  moments  at  the  ends  are  known 
to  be  zero,  and  the  shears  coincide  with 
the  reactions  which  are  found  from  the 
law  of  the  lever.  The  simple  truss  is 
thus  but  a  particular  case  of  the  continu 
ous  one  as  may  be  readily  seen  by  placing 
5=1  in  our  formulae  (I)  to  (IV).  In  an 
end  span  of  a  continuous  truss  the  mo- 


109 

ment  at  the  abutment  is  also  zero  and 
the  shear  is  the  same  as  the  reaction.* 

GIRDERS    WITH    SPANS  ALL    EQUAL. 

This  is  one  of  the  most  common  cases. 
Making  in  (II)  all  the  Ps  equal,  we  have 

c^  —  d^~  0 
c,=dt=  1 

C3  —  "  i =  —  4 
C4=<=  15 
cs=rf.=  -56 

c6=d9=  209,  etc., 

which  are  the  well-known  Clapeyronian 
numbers  first  deduced  by  the  discoverer 
of  the  theorem  of  three  moments. f  Each 
of  these  numbers  is  equal  to  four  times 
the  preceding  one  less  the  one  next  pre 
ceding,  and  their  signs  are  alternately 
positive  and  negative.  They  are  here 
seen  to  be  a  particular  case  of  our  gener 
al  formulae  (II). 

*  For  an  example  of  the  computation  of  a  continuous 
truss  of  two  spans,  see  Van  Nostrand's  Engineering 
Magazine  for  July,  1875. 

t  See  Comptes  JKendus,  1857,  p.  1076. 


110 


GIRDERS  WITH  SYMMETRICAL  SPANS. 

If  the  two  end  spans  of  the  bridge  are 
each  equal  to  ft  I  and  the  others  each 
equal  to  I,  the  multipliers  c  and  d  become 
also  equal.  Their  values  are 


cb—d^—  —  26  --  30  /3 
cG—d^     97  +  112  /?,  etc., 

each  being  equal  to  four  times  the  pre 
ceding  one  less  the  one  preceding  that. 
Having  established  the  value  of  ft  (usual 
ly  taken  at  about  0.8)  these  reduce  at 
once  to  numerical  form.  If  ft  be  unity 
the  spans  become  all  equal  and  the  mul 
tipliers  reduce  to  the  Clapeyronian  num 
bers. 

Example.  —  A  girder  of  four  spans  has 
a  single  load  P  in  the  second^  span  at  a 
distance  k  I  from  the  second  support  ; 
the  two  end  spans  being  equal  to  0.8/ 
and  the  central  ones  to  I.  Find  the  mo 
ment  at  the  second  support. 

Ans.  M  =  P/0.52*—  0.901  7 


Ill 

CONSTANT  AND    VARIABLE  CKOSS    SECTION. 

Having  computed  the  maximum  strains 
in  a  continuous  truss,  we  choose  for  the 
various  pieces  cross  sections  of  an  area 
and  form  sufficient  to  resist  those  strains, 
thus  making  the  girder  one  of  uniform 
strength.  The  theory  by  which  we  have 
computed  the  strains  supposes  however 
that  the  cross  section  of  the  girder  is 
constant.  The  question  now  arises  what 
error  is  introduced  by  this  hypothesis. 

As  we  have  been  unable  to  present  in 
the  short  limits  of  this  paper  the  theory 
of  the  continuous  girder  with  variable 
cross  section,  we  cannot  place  before  the 
reader  a  mathematical  comparison  of  the 
two  cases,  and  are  hence  obliged  to  re 
ly  on  the  computations  of  others  and  on 
general  considerations. 

Computations  of  strains  in  continuous 
girders  have  been  made  by  Bresse,  Mohr, 
Winkler,  Weyrauch  and  others,  consid 
ering  the  cross  section  both  constant  and 
variable.  The  general  conclusion  to  be 
derived  from  their  investigations  is,  that 


112 

the  maximum  moments  over  the  supports 
are  greater,  when  the  variable  cross  sec 
tion  is  taken  into  account,  but  rarely 
more  than  six  per  cent,  that  the  maximum 
moments  near  the  centers  of  the  spans 
are  generally  slightly  less,  and  that  the 
shearing  forces  do  not  sensibly  differ.* 
For  example,  in  a  truss  of  two  equal 
spans,  the  maximum  moment  at  the  pier 
is  0.125  wl*  for  constant  cross  section 
and  0.133  wl*  for  variable  ;  the  maximum 
negative  moment  is  0.070  wl*  for  constant 
and  0.067  w  I*  for  variable,  and  the  reac 
tions  of  the  pier  are  1.250  iv  I  and  1.266 
w  I  respectively. 

If  then  we  compute  continuous  trusses 
as  if  they  were  of  constant  cross  section, 
we  are  liable  to  slight  errors  in  the  chord 
strains.  These  strains  are  however  com 
puted  on  the  assumption  of  a  distribution 
of  live  load  which  can  never  occur  in 
practice,  and  in  proportioning  the  sec- 


*  Mohr.  in  Zeitschrift  des  Arch.  u.  Ing.  Ver.  zu  Hanno 
ver,  1860,  1862,  Winkler ;  Die  Lehre  von  der  Elasticitdt, 
p.  150.  Weyrauch ;  Theorie  der  continuirlichen  Trdger, 
p.  22.,  p.  143. 


113 

tions  to  those  strains  a  factor  of  safety 
involving  five  or  six-fold  security  is  in 
troduced.  Considering  then  that  it 
must  be  almost  impossible  for  the  live 
load  to  be  arranged  on  the  bridge  as 
Fig.  8  represents,  we  may  be  well  as 
sured  that  our  computations  on  the  hy 
pothesis  of  constant  moment  of  inertia 
give  greater  strains  than  can  ever  obtain. 
After  having  computed  on  both  hypo 
theses  a  girder  with  four  spans,  two  of 
sixty-five  meters  in  length  and  two  of 
fifty-two  meters,  Weyrauch  says  :  "  We 
are  now  able  to  answer  the  question, 
whether  it  is  allowable  to  calculate  con 
tinuous  girders  with  variable  cross  sec 
tion  by  the  formulae  for  constant  cross 
section,  in  the  affirmative.  The  maxi 
mum  moments  arising  from  the  two  cal 
culations  differ  but  slightly.  In  our  ex 
ample,  where  the  cross  sections  vary  be- 
twTeen  1  and  24,  the  greatest  difference 
was  6  per  cent,  the  next  following  only 
2.7  per  cent.  The  shears  change  scarcely 
at  all.  Only  for  bridges  with  extremely 
long  spans,  is  it  desirable  to  make  a  sec- 


114 

ond  computation  on  account  of  variable 
cross  section." 


LEVELS  OF  SUPPORTS. 

In  Chapter  I  we  alluded  to  the  fact 
that  small  changes  in  the  relative  levels 
of  the  piers  produce  great  variations  in 
the  strains  of  the  pieces  of  the  truss. 
Continuous  girders  should  not  for  this 
reason  be  used  when  the  piers  are  liable 
to  settle.  Whether  the  supporting  points 
are  exactly  upon  level  when  the  bridge 
is  built  is  a  matter  of  no  importance,  al 
though,  of  course,  no  great  differences 
can  be  allowed. 

If  in  finding  the  equation  of  the  elas 
tic  line  we  had  considered  the  supports 
on  different  levels,  a  term  containing 
those  differences  of  level  and  the  term 
El  would  have  .entered  the  theorem  of 
three  moments.  Now  if  a  straight  beam 
be  laid  across  two  points,  a  downward 
force  is  necessary  in  order  that  it  should 
touch  a  third  point  at  a  lower  level.  If 
this  downward  force  be  furnished  by  the 


115 

weight  of  the  beam,  a  certain  part  of 
this  weight  will  be  effective  in  producing 
the  deflection  or  satisfying  the  term  E  I, 
while  the  remaining  part  will  act  exactly 
as  if  the  three  supports  were  on  the  same 
level.  But  if  the  beam,  instead  of  being 
originally  straight,  were  of  such  a  shape 
that  it  exactly  fitted  the  three  points  it 
is  in  the  same  condition  as  the  horizontal 
one  after  it  has  undergone  the  deflection. 
Hence  its  action  is  independent  of  the 
variations  in  level,  for  no  exterior  force 
is  required  to  compel  it  to  correspond  with 
the  points  of  support. 

We  therefore  conclude,  that  if  a  bridge 
be  built,  by  suitably  adjusting  its  false 
works,  corresponding  to  the  profile  of  the 
piers,  all  the  strains  obtain  exactly  as  if 
those  piers  were  on  one  and  the  same 
level. 


Except  the  well-known  rule  that  the 
lengths  of  the  spans  should  be  so  adjust 
ed  that  the  cost  of  the  piers  and  super 
structure  may  be  equal,  there  is  little  to 


116 

be  said  upon  this  subject.  Although 
most  writers  give  a  mathematical  discus 
sion  of  the  most  economical  relations  of 
spans,  the  fact  that  no  two  of  them 
agree  except  in  the  simplest  case,  only 
indicates  that  the  theory  contains  no 
principle  which  will  lead  to  general  con 
clusions. 

Viewing  the  matter  from  a  practical 
point  of  view,  but  not  neglecting  the  in 
vestigations  of  mathematicians,  we  may 
give  the  following  rules.  For  two  spans 
the  lengths  should  be  equal.  For  others 
the  span  should  be  symmetrical,  the  in 
terior  ones  being  equal  and. the  end  ones 
shorter  by  about  one-fifth  or  one- sixth, 
or  making  the  central  ones  each  equal  to 
/,  the  end  ones  should  be  about  f  I  or  I  L 
Such  an  arrangement  equalizes 'the  mo 
ments  due  to  the  dead  load  and  being 
pleasing  to  the  eye,  it  is  advisable  to  re 
gard  it  in  designing  continuous  bridges. 

PRACTICAL    CONSTRUCTION. 

In  the  construction  of  continuous 
bridges,  the  following  points  should  be 
carefully  observed  : 


117 

1.  The  iron  should  be  of   a  uniform 
qnality,  and  have  undergone  as  near  as 
possible  the   same  process  of  manufac 
ture. 

2.  The  truss  should  be  built  with  par 
allel    chords,  each  capable    of   resisting 
either  tension  or  compression  ;  the  web 
bing   should   be    simple.     With    double 
and  triple  systems  of  webbing  the  strains 
cannot  be  accurately  determined. 

3.  Joints  in  the  chords  should  never 
be  made  over  the  piers. 

4.  The  false  works  should  be  so  adjust 
ed  that  the  bridge  may  be  built  with  its 
points  of  support  on  the  same  relative 
level  with  the  actual  bed  plates. 

5.  The  bearing  surface  of  the  bridge 
upon  the  piers  should  be  as  .small  as  pos 
sible   consistent  with    considerations    of 
strength  and   safety,  and   arrangements 
for  longitudinal  variation,  due  to  changes 
in  temperature,  should  be  provided. 

6.  Continuous  girders  cannot  be  used 
if  the  piers  are  liable  to   variations  in 
level  after  erection. 


118 

ADVANTAGES    OF    THE    CONTINUOUS    SYS 
TEM. 

In  favor  of  the  continuous  girder  versus 
the  simple  one,  we  may  mention  : 

1.  Greater  stiffness,   since  the  deflec 
tion  under  a  rolling  load  is  much  less 
than  that  of  independent  simple  spans. 

2.  Ease   of   erection   in    cases  where 
false  works  are  difficult  and  expensive  ; 
the  girder  may  then  be   built  on   shore 
and  pushed  out  over  the  piers. 

3.  Saving  in    material  for   the   piers, 
since  a  less  bearing  surface  is  required 
than  for  two  ends  of  single  span  bridges. 

4.  Saving  in  iron,  amounting  to  from 
twenty  to  forty  per  cent,  over  the  ordinary 
construction  of  single  spans. 

5.  Simplicity  of  construction,  when  an 
angle  of  skew  exists  in  the  piers:  in  such 
cases  the  cross  girders  may  be  placed  at 
right  angles  in  the  continuous  structure, 
and  the   difficulties   of    oblique  connec 
tions  entirely  avoided. 

In  a  simple  girder,  whose  length  is  /, 
and  live   load  per  unit  of  length  w,  the 


119 


maximum  deflection  due  to  the  live  load 

is ^r-f-  In  two  continuous  spans  each 

384  El 

equal  to  /,  the  maximum  deflection,  which 
occurs  when  one  span  is  covered  with 

3.7  iv  r 

the  live  load,  is  -    -^TTJ    or    only   three- 
384  xL  1 

fourths  as  much.  With  many  continu 
ous  spans,  the  deflection  is  much  less,  its 
greatest  value  being  in  the  end  spans. 
In  the  case  of  a  girder  with  horizontally 
fastened  ends,  the  deflection  is  only  one- 
fifth  of  that  of  ordinary  simple  spans. 

The  saving  in  iron  is  large,  and  alone 
sufficient  to  recommend  the  continuous 
system,  particularly  for  long  spans.  This 
saving  occurs  wholly  in  the  chords  where 
material  can  best  be  spared.  In  the  web 
bing  the  quantity  of  material  is  slightly 
increased.  The  exact  percentage  of  sav 
ing  depends  upon  the  number  and  lengths 
of  spans,  the  proportion  of  live  to  dead 
load,  the  arrangement  of  webbing,  and  will 
be  the  same  in  no  two  cases.  In  the  ex 
ample  of  Fig.  9,  the  center  span  which 
we  computed  does  not  afford  scarcely 


120 

any  saving  in  material  owing  to  the  in 
fluence  of  the  larger  adjacent  spans. 
For  girders  of  two  hundred  feet  in 
length  with  spans  nearly  equal,  calcula 
tion  indicates  a  saving  of  about  thirty 
per  cent.  For  the  extreme  case — a  sin 
gle  span  with  horizontal  fixed  ends — the 
saving  is  fifty  per  cent. 

DISADVANTAGES  OF  THE  CONTINUOUS  SYS 
TEM. 

In  our  last  chapter  we  referred  to  an 
article  by  Charles  Bender,  C.  E.,  which 
contains  many  ingenious  arguments 
against  the  use  of  continuous  bridges. 
There  are,  in  fact,  fifteen  "  conclusions  '' 
to  which  he  is  led  and  which  may  be 
seen  by  the  reader  on  p.  109  of  the  cur 
rent  volume  of  The  Journal  of  the 
American  Society  of  Civil  Engineers. 
Of  these  we  will  give  a  short  abstract 
and  append  a  running  commentary. 

1.  The  theoretical  calculation  of  curves 
of  moments,  without  consideration  of 
proportions  and  details,  is  exceedingly 


121 

fallacious. — Other  things  being  equal, 
the  calculation  of  strains  does  furnish  an 
estimate  of  amounts  of  material.  The 
method  of  curves  moreover  is  neither  so 
accurate  nor  so  quick  as  our  process  of 
tabulation. 

2.  This  fallacy  will  be  greater,  if  the 
theory  stands  upon  false  premises. — Un 
doubtedly. 

3.  The  theory  of  continuity  is  fallaci 
ous  and  unreliable,  because  it  supposes 
the    coefficient    of    elasticity    constant, 
whereas  it  has  been  shown  that  it  may 
vary  for  wrought  iron  from  17,000,000  to 
40,000,000  Ibs.  per  square  inch. — In  our 
last  chapter   we  have   shown   that   the 
proper  interpretation  of  the  variability 
of  E  is,  that  different   qualities  of  iron 
have  different  degrees  of  elasticity.    The 
values  of  E,  from  which  it  is  concluded 
that  the  laws  of  flexure  are  fallacious, 
were  in  fact  found  by  the  theory  itself 
from  the  measured  deflections  of  beams, 
and  it  is  hence  more  than  fallacious  to 
regard  them  as  condemning  that  theory. 

4.  With  several  diagonal  systems  the 


122 

strains  in  a  continuous  girder  cannot  be 
calculated,  but  only  guessed. — A  good 
objection  and  it  applies  also  to  simple 
girders  and  draw  spans. 

5.  The  theory  needs  a  correction  if  the 
chords  are  made  variable  in  cross  section. 
—The  amount  of  this  correction  we  have 
indicated  above. 

6.  The     calculation     of     continuous 
trusses  is  excedingly  tedious,  and  if  gen 
erally  introduced  would  greatly  impede 
the   business  of  bridge  building  in  this 
country. — A  conclusion  to  which  those 
who  know  how  to  calculate  decidedly  ob 
ject.     For  a  construction  costing  half  a 
million  dollars,  it  matters  little  whether 
one  day  or  one  week  be  spent  in  compu 
tation,  and  if  the  one  week  saves  ten  per 
cent  or  more  on  the  cost,  it  is  certainly 
well  employed.* 

*  In  the  discussion  of  this  subject  at  the  late  Engineers1 
Convention,  one  of  the  speakers  mentioned  an  instance 
where  the  strain  sheet  for  a  continuous  revolving  draw 
bridge  was  furnished  for  $40.  As  that  strain  sheet  was 
made  by  the  author  of  this  paper  it  is  not  improper  that 
he  should  remark  that  its  price  would  have  been  consid 
erably  less  bad  the  computations  been  made  by  tabula 
tion  of  apex  loads,  instead  of  by  the  consideration  of 
cases  of  loading. 


123 

7.  Pieces  which    resist  two  kinds  of 
strain  must  be  proportioned  to  resist  the 
maximum   tension    plus    the   maximum 
compression. — Further   experiments  are 
much    needed    in    this    connection  ;    if 
Woehler's     conclusions     are    confirmed, 
pieces  must  be  so  proportioned  and  hence 
the  percentage  of  saving  lowered. 

8.  Continuous   girders    require    very 
accurate  workmanship. — Are  we  to  infer 
that  ordinary  trusses  do  not  ? 

9.  The   foundations   and   masonry  of 
the  piers  must  be  of  excellent  quality.— 
Does  not  the  same  hold  for  the  simple 
girder?     See  No.  11. 

10.  If  the  girders  are  built   on  shore 
and  pushed  out  over  the  piers,  additional 
computations  must  be  made   and  extra 
pieces  introduced. — The  additional  com 
putations  are  of  the   simplest  character 
and  in  many  cases  no  extra  pieces  would 
be  needed. 

11.  If  improperly  placed  on  their  bed 
plates,  greater  strains  arise  than  are  con 
templated,    an    inch    difference  in  level 
producing  great  variations  in  strains. — 


124 

We  have  shown  above  that  unless  piers 
are  liable  to  settle  after  the  erection  of 
the  bridge,  such  differences  in  level  pro 
duce  no  effect. 

12.  If  one    chord    be   protected  from 
the  heat  of  the  sun  the  strains  are  much 
disturbed. — The  camber   of    the  bridge 
is  altered,  as  also  occurs  in  single  spans. 
It  should  be  remembered  however  that 
although  certain  strains  cause  a  curva 
ture,  a  certain  curvature  does  not  neces 
sarily  cause  corresponding  strains. 

13.  Continuous   bridges  have  proved 
to  be  more  economical  in  Europe  than 
simple    spans,    because   the   latter  have 
been  improperly  proportioned. — Trial  is 
necessary  to  prove  that  they  would  or 
would  not  be  more  economical  in  this 
country. 

14.  By  designing  two  bridges  of  200 
feet  each,  a  two  span  continuous  truss 
twenty-five  feet  high,  and  a  simple  truss 
27  feet  high,  with  different  details,  Mr. 
Bender   finds    that  the  latter    is    more 
economical — Perhaps  for  other  propor 
tions  this  might  be  reversed.     In  a  com- 


125 


parison  of  amount  of  material,  ought  not 
other  things  to  be  taken  equal  ? 

15.  Continuous  bridges  deflect  as  much 
as  single  spans. — Not  if  they  have  the 
same  height  and  span,  and  are  subject  to 
the  same  loads  ;  a  fact  which  every 
schoolboy  knows. 

In  regard  to  objections  2,  6,  7,  8,  9, 10, 
12  and  13  one  remark  further  is  neces 
sary.  They  are  objections  which  may 
be  made  to  every  new  proposed  con 
struction  in  engineering  art.  In  the 
days  of  wooden  bridges,  they  were  ad 
vanced  against  the  use  of  iron  ;  they 
have  been  made  against  the  suspension 
system  and  against  the  braced  arch. 
But  their  value  can  be  estimated  in  only 
one  way,  by  trial.  On  the  other  hand 
theory  can  estimate  one  at  least  of  the 
advantages  claimed  for  the  continuous 
systems,  and  that  estimate  is  a  saving  of 
twenty  to  forty  per  cent,  in  material; 
how  much  of  this  must  be  deducted  for 
extra  care  in  workmanship,  labor  of  erec 
tion,  effects  of  temperature,  or  varia 
tions  in  the  elasticity  of  the  material  can 


126 

only  be  determined  by  the  actual  erec 
tion  of  continuous  bridges,  by  experi 
ments  extending  through  a  long  series 
of  years. 

mr 

Having  thus  stated  briefly,  but  fairly, 
the  arguments  for  and  against  the  use  of 
continuous  bridges,  we  leave  it  to  prac 
tical  builders  to  decide  whether  or  not 
the  system  is  worth  a  trial.  Other  na 
tions  have  long  been  using  it ;  profiting 
by  their  experience  and  by  our  own  im 
proved  methods  of  manufacture  and 
modes  of  erection,  it  may,  perhaps,  turn 
out  that  we  shall  find  it  better  and  more 
economical  than  the  present  system  of 
single  independent  spans. 

HfSTORY    AND    LITERATURE. 

The  literature  on  the  theory  of  con 
tinuous  girders  is  very  extensive  in  the 
German  and  French  languages,  and  very 
limited  in  English.  We  can  only  give 
here  a  few  hints  concerning  its  develop 
ment  and  history. 

About  the  year  1825,  Navier  founded 
the  present  theory  of  flexure  by  intro- 


127 


ducing  the  hypothesis  that  the  exten 
sions  and  compressions  of  the  fibers  on 
each  side  of  the  neutral  axis  were  pro 
portional  to  their  distances  from  that 
axis.  From  this  he  deduced  the  equa 
tion  of  the  elastic  line,  and  applied  it  to 
the  discussion  of  continuous  girders. 
His  method  consisted  in  determining 
first  the  reactions  of  the  supports,  and 
from  these  the  internal  forces  or  strains, 
which,  although  the  most  logical,  was 
exceedingly  tedious  in  practice.  In  the 
following  works  the  reader  may  find  de 
tailed  information  concerning  his  meth 
od  : 

Kayser  :  Handbuch  der  Statik,  Carls- 
ruhe,  1836,  chap.  X. 

Molinos  et  Pronnier  :  Traite  de  la 
construction  des  ponts  metaliques,  I^aris, 
1857. 

From  the  time  of  Navier  to  the  pub 
lication  in  1857  by  Clapeyron  of  the 
method  of  using  the  moments  over  the 
piers  as  auxiliaries  in  the  computation 
instead  of  the  reactions,  many  continu 
ous  girders  were  built  in  France,  Ger- 


128 


many  and  England.  Among  these  may 
be  mentioned  the  Britannia  tubular 
bridge,  of  four  spans,  two  of  231  feet, 
and  two  of  460  feet  ;  the  Boyne  Via 
duct  with  three  spans  of  141,  267  and 
141  feet ;  and  the  bridge  over  the 
Weichsel  at  Dirschau  with  six  continu 
ous  spans,  each  397  feet  in  length.  By 
Clapeyron's  happy  discovery  of  the  the 
orem  of  three  moments,  a  great  impetus 
seemed  to  be  given  toward  the  erection 
of  such  bridges,  for  in  the  twenty  years 
following  1857,  we  find  them  extensively 
built  in  Germany  and  France,  although 
in  England  the  unfortunate  example  of 
the  Britannia  tube  caused  a  tendency  to 
other  forms  of  construction.  A  mere 
list  of  such  bridges  would  occupy  pages. 
They  are  generally  of  shorter  span  than 
those  mentioned  above,  rarely  exceeding 
300  feet,  while  the  number  of  spans 
varies  from  two  to  seven. 

To  give  here  a  list  of  the  books  which 
has  appeared  since  Clapeyron's  time  will 
also  be  impossible.  We  can  only  indi 
cate  two  or  three  which  are  at  the  same 
time  valuable  and  easily  accessible  : 


129 

Bresse :  La  mecanique  appliqu&e, 
Paris,  1865.  (Vol.  Ill  contains  a  tolera 
bly  complete  mathematical  discussion, 
with  tables  for  facilitating  the  calcula 
tion  of  moments.) 

Winkler:  Theorie  der  J5rM6c&6?i,  Vienna, 
1872.  (Contains  the  graphical  method 
of  Culmann  and  Mohr,  with  also  analyt 
ical  investigations.) 

Weyrauch  :  Theorie  der  continuir- 
lichen  Traeger^  Leipzig,  1873.  (A  com 
plete  analytical  discussion  of  the  whole 
subject.  The  best  work  which  has  yet 
appeared.) 

In  our  own  country  have  been  issued 
during  the  past  year  the  works  of 
Greene,  Herschel  and  DuEois,  each  of 
which  contains  valuable  contributions  to 
the  literature  of  the  subject,  and  which 
should  be  in  the  library  of  every  student 
of  engineering.  On  the  other  hand, 
works  by  English  authors,  which  treat  of 
the  subject  at  all,  do  it  in  such  an  imper 
fect  and  unsatisfactory  manner,  that  we 
are  forced  to  consider  them  as  twenty 
years  behind  the  age. 


130 

As  the  above  mentioned  works  treat 
only  of  the  theory  and  calculation  of 
continuous  girders,  we  ought  perhaps  to 
say  that  a  book  giving  an  account  of  the 
most  important  continuous  and  simple, 
together  with  suspension  and  arch 
bridges,  is  the  admirable  descriptive 
work  by  Heinzerling,  Die  BruecJcen  in 
Eisen:  Leipzig,  1870,  which  is  illustrat 
ed  by  over  a  thousand  engravings,  and 
presents  a  complete  history  of  iron 
bridge  construction. 


ERRATA. 

Page  15,  line  13  ;  for  ^  read  I. 
11     21,     "17  and  18  ;  for  wl9  read  wl*. 


14  DAY  USE 

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